KOH. Option B is correct
Explanations:A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.
The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)
4. An ice cube (25 g) is at -8.0°C. How much energy is required to take it to the
melting point, 0 °C? Heat capacity (c) for solid water is 2.10 J/g C
the heat energy required to take it to the melting point is 420J.
What is specific heat capacity?The heat capacity, abbreviated Cp, is the amount of heat needed to elevate a mole of a substance's heat content by precisely one degree Celsius.
A material has more thermal energy the hotter it is, says basic thermodynamics. Additionally, when a chemical is present in greater concentrations at a particular temperature, it will have a higher total thermal energy.
Mathematically,
Q = mc∆T
Where,
Q = quantity of heat absorbed by a body
m = mass of the body
∆t = Rise in temperature
C = Specific heat capacity of a substance depends on the nature of the material of the substance.
S.I unit of specific heat is J kg-1 K-1.
Given,
An ice cube of mass = m=25g
initial temperature = T1 = -8°C
final temperature = T2 = 0°C
Heat capacity for solid water = c = 2.1J/g°C
According to heat energy required to take it to the melting point,
Q = mc∆T
Q = 25g×2.1J/g°C × (0+8) °C
Q = (25×2.1×8) J
Q = 420J
Hence, the heat energy required to take it to the melting point is 420J.
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A compound containing 63.88% Cl and 36.12% Ca has a molecular mass of 443.92 g/mol, what is the molecular formula?
The molecular formula of a compound containing 63.88% Cl and 36.12% Ca that has a molecular mass of 443.92 g/mol is Ca₄Cl₈.
How to calculate molecular formula?Molecular formula is the notation indicating the number of atoms of each element present in a compound.
To calculate the molecular formula of a compound, the empirical formula must first be calculated as follows:
63.88% Cl = 63.88g ÷ 35.5g/mol = 1.79mol36.12% Ca = 36.12g ÷ 40g/mol = 0.903molCl = 1.79mol ÷ 0.903 = 1.98Ca = 0.903mol ÷ 0.903 = 1The approximate empirical ratio of Cl and Ca is 2:1, hence, the empirical formula is CaCl₂.
{CaCl₂}n = 443.92
{(40 + 35.5(2)}n = 443.92
111n = 443.92
n = 4
The molecular formula is Ca₄Cl₈.
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Answer:
Ca4Cl8
Explanation:
Canva
2022
Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled 24 ∘C at a pressure of 745 mmHg . A balloon with a volume of 443 mL drifts into the airlock where the temperature is − 95 ∘C and the pressure is 0.115 atm . What is the final volume, in milliliters, of the balloon if n does not change and the balloon is very elastic?
Answer: the final volume of the balloon is 2.26 x 10^3 mL
Explanation:
The question requires us to determine the new volume of a balloon, given the initial and final conditions.
The following information was provided by the question:
initial temperature = T1 = 24 °C = 297.15 K
initial volume = V1 = 443 mL
initial pressure = P1 = 745 mmHg = 0.980 atm
final temperature = T2 = -95 °C = 178.15 K
final pressure = P2 = 0.115 atm
To solve this problem, we'll need to apply the equation of ideal gases to calculate the number of moles of gas in the balloon, and then use this value and the final temperature and pressure provided to determine the final volume,
The equation of ideal gases can be written as:
[tex]PV=nRT[/tex]And we can rearrange it to calculate the number of moles:
[tex]PV=nRT\rightarrow n=\frac{PV}{RT}[/tex]Applying the values provided by the question:
[tex]n=\frac{(0.980atm)\times(443mL)}{R\times(297.15K)}=\frac{1.46}{R}(\frac{atm\times mL}{K})[/tex]Now, we can rearrange the equation of ideal gases to calculate the volume:
[tex]PV=nRT\rightarrow V=\frac{nRT}{P}[/tex]And, applying the values provided and the number of moles as calculated:
[tex]V=\frac{(\frac{1.46}{R}atm.mL.K^{-1})\times R\times(178.15K)}{0.115atm}=2.26\times10^3mL[/tex]Therefore, the final volume of the balloon is 2.26 x 10^3 mL.
SYNTHESIS OFCARBONATECTIONLABORATORY SIMULATIONLab Data- X99.00.10CollectedVolume sodium carbonate (mL)Molarity sodium carbonate (M)Volume calcium chloride (mL)Molarity calcium chloride (M)ObservationsThe mixture has now turned white100.00.20hemDisp0.211.30Mass filter paper (9)Mass filter paper + precipitate (9)CalculatedObserved mass calcium carbonate (9)Identify limiting reactantExpected mass calcium carbonate (9)Percent yield (%)0.88Calcium chlorideHow to calculate theoretical yield and percent yield
Answer:
[tex]\begin{gathered} \text{Limiting Reagent = Sodium Carbonate} \\ \text{Percent Yield = 98\%} \end{gathered}[/tex]Explanation:
The chemical reaction talks about the synthesis of calcium carbonate
It is from the reaction between sodium carbonate and calcium chloride
Let us write the equation of reaction as follows:
[tex]Na_2CO_{3(aq)}+CaCl_{2(aq)}\text{ }\rightarrow2NaCl_{(s)\text{ }}+CaCO_{3(aq)}[/tex]Firstly, we want to get the expected mass of calcium carbonate
This speaks about getting the theoretical yield based on the equation of reaction
From the data collected, 90 ml of 0.20 M (mol/L) of sodium carbonate gave calcium carbonate
We need to get the actual number of moles of sodium carbonate that reacted
We can get this by multiplying the volume by the molarity (kindly note that we have to convert the volume to Liters by dividing by 1000)
Thus, we have it as:
[tex]\frac{90}{1000}\times\text{ 0.1 = 0.009 moles}[/tex]Hence, we see that 0.009 moles of sodium carbonate reacted theoretically
Since 1 mole of sodium carbonate gave 1 mole calcium carbonate, it is expected that 0.009 mole of sodium carbonate will give 0.009mole of calcium carbonate
What we have to do now is to get the theoretical grams of calcium carbonate produced
That would be the product of the number of moles of calcium carbonate and its molar mass
The molar mass of calcium carbonate is 100 g/mol
The theoretical yield (expected mass) is thus:
[tex]100\text{ g/ mol }\times\text{ 0.009mol = 0.9 g}[/tex]Finally, we proceed to get the percentage yield which is calculated using the formula below:
[tex]\text{Percent Yield = }\frac{Actual\text{ yield}}{\text{Theoretical yield}}\times\text{ 100 \%}[/tex]The actual yield is the observed mass which is given as 0.88 g
The percent yield is thus:
[tex]\frac{0.88}{0.9}\times\text{ 100 = }98\text{ \%}[/tex]
A 7.94 g of solid CO2 (Dry ice) is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 301 K. What is the pressure?
In order to find the pressure in this situation, we will be using the Ideal gas Law, which perfectly correlates these informations, the formula for this Law is:
PV = nRT
Where:
P = pressure, we want to find it
V = volume, 1.00 L
n = number of moles, we will also find it
R = is the gas constant, 0.082
T = 301 K
To find the number of moles, we need to use the mass provided in the question and also the molar mass of CO2, 44.01g/mol
44.01g = 1 mol
7.94g = x moles
x = 0.180 moles of CO2
Now we can use the ideal gas formula:
P * 1.00 = 0.180 * 0.082 * 301
P = 4.44 atm
A quantity of 3.30 × 102 mL of 0.500 M HNO3 is mixed with 3.30 × 102 mL of 0.250 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46°C. The heat of neutralization when 1.00 mol of HNO3 reacts with 0.500 mol Ba(OH)2 is −56.2 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively). What is the final temperature of the solution?
The final temperature of the solution is 20 °C.
Given that :
volume of HNO₃ = 0.33 L
molarity = 0.500 M
no. of moles HNO₃ = 0.33 × 0.500
= 0.165 mol
moles of Ba(OH)₂ = 0.33 × 0.250
= 0.082 mol
Q = ΔH × n
Q = - 52600 J/mol × 0.165 mol
Q = - 8679 J
total volume = 330 mL +330 mL = 660 mL
Density = mass / volume
mass = 1 g/mL × 660 mL
mass = 660 g/mL
now, the change in temperature can be calculated by following formula :
Q = mc ( ΔT )
Q = mc ( T2 - T1 )
- 8679 = 660× 4.184 ( T2 - 18.46 °C)
T2 = 20 °C
Thus, A quantity of 3.30 × 10² mL of 0.500 M HNO3 is mixed with 3.30 × 10² mL of 0.250 M Ba(OH)₂ in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46°C. The heat of neutralization when 1.00 mol of HNO₃ reacts with 0.500 mol Ba(OH)₂ is −56.2 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively). the final temperature of the solution is 20 °C.
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when a bunsen burner is properly adjusted, what should the flame look like?
when a bunsen burner is properly adjusted, the flame looks Blue with no gap between the burner and the flame.
A Bunsen burner is a particular kind of gas burner that is frequently utilized as a heat source in lab investigations. The barrel or chimney of the burner is made up of a straight tube that extends vertically from a flat base. At the bottom of the chimney, liquid petroleum gas, such as propane or butane, or natural gas, primarily methane, is delivered.
The base of the chimney on Bunsen burners typically has a hose barb installed so that rubber tubing may supply the gas from a gas nozzle on the lab bench.
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What is the correct formula that wouldresult from the combination of the twoionic species?Na¹+ and CIO ¹-1-4BNa₂(CIO4)2NaCIO4
To determine how the formula will be when combining two ionic species, we must look at the oxidation states of the ions. The Na ion has an oxidation state of +1 and the ClO4 ion has an oxidation state of -1.
The number of ions must be such that the total sum of the oxidation states is zero. If we have a Na+1 ion and a ClO4(-1) ion, the sum will be equal to zero. So there must be one ion of each species in the molecule.
[tex]Na^{+1}ClO_4^{-1}\rightarrow NaClO_4[/tex]Therefore, the answer will be: NaCIO4
. At standard pressure, what state of matter is xenon at -111 °C?
A) solid
B) liquid
C) gas
D) can be both solid and liquid
E) can be both liquid and gas
At standard pressure, C) gas state of matter is xenon at -111 °C
What is the state of matter of xenon?Xenon is an extremely uncommon, odourless, colourless, tasteless, chemically inert gas. It was thought to be absolutely innocuous until Neil Bartlett published the synthesis of xenon haxafluoroplatinate in 1962. When stimulated by an electrical discharge, xenon generates blue light in a gas-filled tube.
Noble gases, which are most commonly encountered as monatomic gases, have entirely filled outer electron shells and hence have little desire to react with other elements, resulting in relatively few compounds with other elements.
In its +6 oxidation state, xenon trioxide is an unstable molecule. It is a highly potent oxidising agent that progressively liberates oxygen from water when exposed to sunshine. When it comes into touch with organic materials, it becomes highly explosive.
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PLEASE HELP, WILL MARK BRANLIEST!!
The half-life of cobalt-60 is 10.47 min. How many grams of cobalt-60 remain after 104.7 min if you start with 1024g?
Answer: 1 gram
Explanation:
- the equation is 1024(.5)^10
• so 1024g is your initial amount and it’s the amount that is being decayed
• since it’s half-life, the decay factor is 0.5
• the half life is 10.47 minutes, so 10.47 minutes would be considered one round of half life.
-basically, if you were finding the half life after 10.47 minutes you would use the equation 1024(.5)^1.
-as said, it’s raised to the 1st power because 10.47 minutes is one round.
• since you’re looking for the amount after 104.7 minutes, the equation is raised to the 10th power because 104.7 / 10 = 10.47. so 104.7 is 10 rounds of half-life.
- so 1024(.5)^10 = 1
so the half-life of 1024g of Cobalt-60 after 104.7 minutes is 1
hope this helps :)
I need help on balancing the equations and on what type of reaction it is.
Given the balanced chemical reaction expressed as:
[tex]2Au_2O_3\rightarrow4Au+3O_2[/tex]A Redox reaction is a reaction that involves the transfer of electrons and changes in the oxidation state between the elements.
The given chemical equation is therefore an oxidation-reduction reaction since it involves a change in the oxidation state of the elements.
Determine the moles of Au₂O₃
[tex]\begin{gathered} \text{Mole = }\frac{Mass}{Molar\text{ mass}} \\ \text{Mole of Au}_2O_3=\frac{10g}{441.93g\text{/mol}} \\ \text{Mole of Au}_2O_3=0.02263\text{moles} \end{gathered}[/tex]According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 4 moles of Gold. Hence the moles of Gold produced will be:
[tex]\begin{gathered} \text{moles of Gold=}\frac{0.02263\times4}{2} \\ \text{moles of Gold=}0.02263\times2 \\ \text{moles of Gold=}0.0453\text{moles} \end{gathered}[/tex]Determine the mass of Gold.
[tex]\begin{gathered} \text{Mass of Gold=moles}\times molar\text{ mass} \\ \text{Mass of Gold=}0.0453\times196.97 \\ \text{Mass of Gold}=8.91\text{grams} \end{gathered}[/tex]Next is determining the mole of Oxygen
According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 3 moles of Oxygen. Hence the moles of Oxygen produced will be:
[tex]\begin{gathered} \text{moles of Oxygen=}\frac{0.02263\times3}{2} \\ \text{moles of Oxygen}=0.033945\text{moles} \end{gathered}[/tex]Determine the mass of oxygen produced
[tex]\begin{gathered} \text{Mass of O}_2=moles\times\text{Molar mass} \\ \text{Mass of O}_2=0.033945\times16 \\ \text{Mass of O}_2=0.543\text{grams} \end{gathered}[/tex]Hence the mass of oxygen produced is 0.543 grams
How many significant figures (SF) are in each of the following measured quantities? Drag the appropriate measurement to the respectable bins
Chemistry => Measurements => Significant Figures
Significant figures correspond to the number of digits in a number. We have to take into account the following:
• Zeros at the beginning of a number or at the end are not counted as digits, but zeros in between the number should be counted.
,• When we write a number in scientific notation, the 10 that accompanies the number should not be counted.
Following these indications, let's count the significant figures of each given number:
In summary, we have the answer will be:
1SFs
3.00 m
4.0x10^3 mL
3SFs
50 100 00g
80.10 mL
60.4 °C
6SFs
9018.17 kg
Calculate the pH of the resulting solution when 25.0mL of 0.30M HClO4 is added to 60.0ml of 0.35 M CH3NH2. Kb for CH3NH2 = 4.4*10^-4
The pH of the resulting solution is 10.9 .
given that :
for HClO₄
volume = 25 mL = 0.025 L
M = 0.30 M
no. of moles = 0.025 × 0.30
= 0.0075
for CH₃NH₂
no. of moles = 0.35 × 0.06
= 0.021
as a resultant no. of moles of CH₃NH₂ that remain in solution
= 0.021-0.0075
= 0.0135
total volume = 0.025 + 0.060
= 0.085
the concentration will be :
concentration for HClO₄ = 0.0075 / 0.085
= 0.088 M
concentration for CH₃NH₂ = 0.0135 / 0.088
= 0.153
kb = 4.4 × 10⁻⁴
pkb = - log kb
pkb = 3.3
the pOH formula is given as :
POH = pKb + log [ acid ] / [ base ]
pOH = 3.3 + log [ 0.088 ] / [ 0.153 ]
pOH = 3.06
pH + pOH = 14
pH = 14 - 3.06
pH = 10.9
Thus, The pH of the resulting solution when 25.0mL of 0.30M HClO₄ is added to 60.0ml of 0.35 M CH₃NH₂. Kb forCH₃NH₂ = 4.4 × 10^-4 . pH is 10.9.
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PLS HELP, IT’S DUE TODAY
In a nuclear fusion reaction, two small, light ___ (hydrogen atoms) combine under extreme __ and pressure to form one larger, heavier nucleus (helium).
Answer:
In a nuclear fusion reaction, two small, light 'nuclei' (hydrogen atoms) combine under extreme 'temperature' and pressure to form one larger, heavier nucleus (helium).
An experiment was designed to test the hypothesis that peanuts have more energy than a chip. The experiment determines calorimetry of the peanut and the chip, using water to capture the energy of the samples when they were burned. A 10 g sample of chip and a 15 g sample of peanut was burned underneath a metal can that held 50 g of water. A thermometer captured how much the temperature of the water increased. The water increased by 7 degrees for the peanut and 3 degrees for the chip. What error was made?
a The temperatures were read wrong.
b Different amounts of food samples were used.
c The food samples were too similar.
d Different amounts of water were used.
Answer:
b Different amounts of food samples were used.
Explanation:
The mass of the two samples needs to be the same in order for the test to be accurate.
How many moles of argon atoms are present in 11.2 L of argon gas at STP? Round your molar mass to whole numbers andgive your answer in the correct number of sig figs and units
STP or Standard Pressure and Temperature is widely used when the subject is gases. The used standard temperature is 273 K or 0°C and the standard temperature used is 1 atm. At these conditions 1 mol will be equal to 22.4 L of volume, therefore if we have 11.2 L of Argon:
22.4 L = 1 mol
11.2 L = x moles of Argon
22.4x = 11.2 L
x = 0.500 moles of Ar will be present in 11.2 L of volume
PLEASE HELP!!
For this car, the airbag must have a volume of 58 liters when fully inflated. To provide an adequate cushion for the driver’s head, the air pressure inside the airbag should be 4.4 psi. This pressure value is in addition to the normal atmospheric pressure of 14.7 psi, giving a total absolute pressure of 19.1 psi, which equals 1.30 atmospheres.
One of the main components of an airbag is the gas that fills it. As part of the design process, you need to determine the exact amount of nitrogen that should be produced. Calculate the number of moles of nitrogen required to fill the airbag. Show your work. Assume that the nitrogen produced by the chemical reaction is at a temperature of 495°C and that nitrogen gas behaves like an ideal gas.
Part C
Recall the balanced chemical equation from part B of task 1:
2NaN3 → 2Na + 3N2.
Calculate the mass of sodium azide required to decompose and produce the number of moles of nitrogen you calculated in part B of this task. Refer to the periodic table to get the atomic weights.
Part D
What would happen if the amount of sodium azide used was far greater or far less than what you calculated in part C? Describe both cases.
The number of moles of nitrogen required to fill the airbag is 1.197 moles.
The mass of sodium azide required to decompose and produce 1.197 moles of nitrogen is 51.87 g.
If the mass used was greater, the airbag could burst, but if the mass used was smaller, the airbag would not inflate properly.
What amount in moles of nitrogen is required to fill the bag?The number of moles of nitrogen required to fill the airbag is calculated from the equation of reaction as follows:
Using the ideal gas equation; PV = nRT
where;
P = pressureV = volumen = number of molesR = molar gas constantT = temperatureFrom the data provided:
P = 1.30 atm
V = 58 Liters
n = ?
R = 0.082 atm.L.mol⁻¹K⁻¹
T = 495 °C or ( 273.15 + 495) K = 768.15 K
solving for n;
n = PV/RT
n = (1.3 * 58) / (0.082 * 768.15)
n = 1.197 moles
Equation of reaction: 2 NaN₃ → 2 Na + 3 N₂
moles ratio = 2 moles of sodium azide produce 3 moles of N₂
Moles of azide required = 1.197 * 2/3 = 0.798 moles
molar mass of sodium azide = 65 g/mol
mass of sodium azide = 0.798 * 65
mass of sodium azide required = 51.87 g
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metal oxide MO2 reacts with excess HCl to produce chlorine gas at STP as given by the following unbalanced equation:MO2 (s) + HCl (aq) -,> MCl2 (aq) + Cl2 (g) + H20 (l) a. determine limiting reactantb. calculate the mass of MCl2 produced in the reactionc. calculate the percentage yield if the actual mass of MCl2 produced is 0.078g
1st) It is necessary to balance the chemical equation:
[tex]MO_2+4\text{HCl}\rightarrow MCl_2+Cl_2+2H_2O[/tex]2nd)
give the n and l values and the number of orbitals for sublevel 5g.
The n and l values and the number of orbitals for sublevel 5g is :
5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.
There are total four quantum numbers:
1) Principal quantum number , n
2) Angular quantum number , l
3) Magnetic quantum number , ml
4) spin quantum number , ms
For 5g shell, n = 5
subshell g , l = 4 ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g
number of orbitals in subshell = (2l + 1) ( 2×4 + 1) = 9
Thus, The n and l values and the number of orbitals for sublevel 5g is :
5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.
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I’m getting certain numbers but I want to be sure I’m doing this right
To write the numbers with the specific number of significant figures, we need to remember that zeros to the left don't count as a significant figure and also that we can add zeros to the right to add significant figures to the numbers.
1) 1.5408 to 3 significant figures, we start counting from left to right, so we will maintain the numbers until the second decimal place, the "8". Since the next is "0", the rounding is "8", so the number is 1.54.
2) Here is similar, but since we don't have decimal places, we need to put the zeros in place of the numbers we are rounding. Since we want 2 significant figures, we will maintain only the first two. However, since the third is greater than 5, we need to round up, so the number is 4400.
3) To get this, we will maintain the zeros to the elft, but they don't count, so we start counting from the "1". Since the third number is less than 5, we round down, so the number is 0.019.
4) Now, here we have an example we need to add zeros. The way it is, 0.5 has only 1 significant figure, so we need to add 3 more zeros to the right to get to 4 significant figures. The number is 0.5000.
5) Here, is the same as item 3, but we want 3 significant figures and since the fourth is greater than 5 we round it up. So, the number is 0.066.
what is the limiting and excess reactant if 15.0g of FePo4 reacts with 5.0g of Na2SO
Explanation:
We are given: mass of FePO4 = 15g
: mass of Na2SO4 = 5g
We first find the mass of Fe2(SO4)3 from the mass of FePO4:
m is the mass and M is the molar mass
[tex]\begin{gathered} m\text{ = }\frac{m(FePO4)}{M(FePO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{2mol\text{ FePO4}}\times\text{ M\lparen Fe2\lparen SO4\rparen3\rparen} \\ \text{ = }\frac{15}{150.82}\times\frac{1}{2}\times399.88 \\ \text{ = 19.89g} \end{gathered}[/tex]We then find the mass of Fe2(SO4)3 from Na2SO4:
[tex]\begin{gathered} m\text{ = }\frac{m(Na2SO4)}{M(Na2SO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{3mol\text{ Na2SO4}}\times M(Fe2(SO4)3) \\ \text{ =}\frac{5}{142.04}\times\frac{1}{3}\times399.88 \\ \text{ = 4.69g} \end{gathered}[/tex]Answer:
Therefore, FePO4 is the excess reactant and Na2SO4 is the limiting reactant
DUE AT 11:59 PLEASE HELP
A student has a calorimeter with 211.7 grams of 20.4 degrees Celsius water contained within it. The student then adds 128.9 grams of 94.2 degrees Celsius water to that calorimeter and stirs. To what maximum temperature will the cold water in the calorimeter rise to?
The maximum temperature the cold water in the calorimeter will rise to is 48.3 °C
How to determine the maximum temperatureThe maximum temperature the cold water can attain can be obtained by calculating the equilibrium temperature. This can be obtained as follow:
From the question given above, the following data were obtained:
Mass of cold water (M) = 211.7 grams Temperature of cold water (T) = 20.4 °CMass of warm water (Mᵥᵥ) = 128.9 gramsTemperature of warm water (Tᵥᵥ) = 94.2 °CSpecific heat capacity of the water (C) = 4.18 J/gºC Equilibrium temperature (Tₑ) =?Heat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = MC(Tₑ – T)
Cancel out C
Mᵥᵥ(Tᵥᵥ – Tₑ) = M(Tₑ – T)
128.9 × (94.2 – Tₑ) = 211.7 × (Tₑ – 20.4)
Clear bracket
12142.38 – 128.9Tₑ = 211.7Tₑ – 4318.68
Collect like terms
12142.38 + 4318.68 = 211.7Tₑ + 128.9Tₑ
16461.06 = 340.6Tₑ
Divide both side by 340.6
Tₑ = 16461.06 / 340.6
Tₑ = 48.3 °C
Thus, we can conclude that the maximum temperature is 48.3 °C
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If 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal?
The specific heat capacity of the metal that needs 1495 J of heat to raise the temperature is 0.43J/g°C.
How to calculate specific heat capacity?Specific heat capacity is the heat capacity per unit mass of a substance. It can be calculated by using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of substancec = specific heat capacity∆T = change in temperatureAccording to this question, 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C. The specific heat capacity can be calculated as follows;
1495 = 315 × c × {66 - 55}
1495 = 3465c
c = 1495/3465
c = 0.43J/g°C
Therefore, 0.43J/g°C is the specific heat capacity of the metal that requires a heat of 1495J.
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How many atoms of oxygen are in the reactants of the following equation? 3NO2 + H2O --> NO + 2HNO3
The reactants are the compounds before the arrow and the products are after the arrow. Just as they show us, the oxygen atoms in the reagents will bJust as the reaction shows, the oxygen atoms in the reactants will be:
Answer: There are 7 atoms of oxygen in the reactants
How many grams of CO2 are produced if 12 liters of H2O are produced?
Answer
[tex]14653.7181\text{ g }CO_2\text{ }[/tex]Procedure
Considering the following balanced equation.
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
To determine the grams of CO₂ that will be produced, you will first need to get the grams of water produced, then convert the grams of water into moles of water and use the molar proportios to get the moles of CO₂. Lastly, convert the moles of CO₂ into grams.
Additional data:
Dnsity of water = 1 g H₂O=/ of H₂O
CO₂ molar mass = 44.01 g/mol
H₂O molar mass = 18.02 g/mol
Calculations
Convert to mass
[tex]12L\text{ }H₂O\text{ }\frac{1kg\text{ }H_2O}{1L\text{ }H_2O}=12kg\text{ }H_2O[/tex]Convert to moles
[tex]12kg\text{ }H₂O\text{ }\frac{1000g\text{ }H_2O\text{ }}{1kg\text{ }H_2O}\frac{\text{ }1\text{ }mole\text{ }H_2O}{18.02g\text{ }H_2O}=665.9267\text{ }mole\text{ }H_2O[/tex]Convert to moles of CO₂ using the relationship
[tex]665.9267\text{ }mole\text{ }H_2O\frac{2moleCO_2}{4mole\text{ }H_2O}=332.9634\text{ }mole\text{ }CO_2[/tex]Convert to grams of CO₂
[tex]332.9634\text{ }mole\text{ }CO_2\text{ }\frac{44.01g\text{ }CO_2}{1\text{ }mole\text{ }CO_2}=14653.7181\text{ g }CO_2\text{ }[/tex]Which naturally occurring gas can be found in certain rocks and soils and is considered a hazardous air pollutant?A. argonB. carbon dioxideC. radonD. mercury
Radon is a naturally-occuring radioactive gas that can cause lung cancer to those who are near it. Radon is also an inert, colorless and odorless gas. Although it is dangerous and hazardous, it disperses rapidly and, usually, is not a health issue. Answer is Radon
An organic compound contains carbon hydrogen and oxygen. If it contains 45.27% carbon and 4.43 % hydrogen by mass determine the empirical formula
This organic compound has in its structure the next element: C, H, and O
45.27% C
4.43 % H
The rest 100%-45.27%-4.43 = 50.3 % O
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The empirical formula of a chemical compound is the simplest whole-number ratio of atoms present in a compound.
Procedure:
Step 1)
We convert % into grams (g). Let's assume we have a 100g sample. Therefore,
45.27 % C = 45.27 g
4.43 % H = 4.43 g
50.3 % O = 50.3 g
Step 2)
We calculate the number of moles of each element. To do this, we need every atomic mass
For C)
[tex]45.27\text{ g x }\frac{1\text{ mol}}{12.01\text{ g}}=3.767\text{ moles}[/tex]For H)
[tex]4.43\text{ g x }\frac{1\text{ mol}}{1.00\text{ g}}=\text{ 4.43 moles}[/tex]For O)
[tex]50.3\text{ g x }\frac{1\text{ mol}}{16.0\text{ g}}=3.14\text{ moles}[/tex]Step 3)
We divide all moles by the smallest one of them (3.14 moles of O)
For C) 3.767 moles/3.14 moles = 1.19 = 1 (we need integer numbers)
For H) 4.43 moles/3.14 moles = 1.41 = 1
For O) 3.14 moles/3.14moles = 1
The empirical formula is CHO
Express the following in liters at STP:
4.83 x 10^-3 moles HF
Considering the definition of STP conditions, 4.83 × 10⁻³ moles of HF will occupy a volume of 0.108192 L at STP.
STP conditionsThe STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases, and 1 mole of any gas occupies an approximate volume of 22.4 liters.
Volume in this caseIn this case, you have 4.83×10⁻³ moles of HF. You can apply the following rule of three: if by definition of STP conditions 1 mole of HF occupies a volume of 22.4 liters, 4.83×10⁻³ moles occupies how much volume?
volume= (4.83×10⁻³ moles ×22.4 L)÷ 1 mole
volume= 0.108192 L
Finally, there is a volume of 0.108192 L at STP.
Learn more about STP conditions:
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A redox reaction:A. is a reaction where oxygen is returned it its natural state.B. is a reduction reaction where oxygen is removed.C. is comprised of two half reactions, a reduction and an oxidation.D. is a reaction where oxygen is added to a compound.
In this question, we have to classify what is a redox reaction, and as the name suggests, we have a reduction and an oxidation reaction occurring in this type of reaction, where one element will lose electrons, or will be oxidized, and another element will gain electrons or will be reduced. Therefore the best answer will be letter C
help me please if you can a. ammonia b. battery acid c.pure waterd. sea water
Answer:
ammonia
Explanation: