At 302 K , to what pressure can the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire? (Note that the gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi .)

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 :)

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).

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

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.

1st) It is necessary to balance the chemical equation:

2nd)

Given the balanced chemical reaction expressed as:

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

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:

Determine the mass of Gold.

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:

Determine the mass of oxygen produced

Hence the mass of oxygen produced is 0.543 grams

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.

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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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.

Therefore, the answer will be: NaCIO4

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

At standard pressure, C) gas state of matter is xenon at -111 °C

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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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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Answer

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

Convert to moles

Convert to moles of CO₂ using the relationship

Convert to grams of CO₂

The specific heat capacity of the metal that needs 1495 J of heat to raise the temperature is 0.43J/g°C.

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;

According 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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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

The maximum temperature the cold water in the calorimeter will rise to is 48.3 °C

The 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:

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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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

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

We then find the mass of Fe2(SO4)3 from Na2SO4:

Answer:

Therefore, FePO4 is the excess reactant and Na2SO4 is the limiting reactant

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

the heat energy required to take it to the melting point is 420J.

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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Answer:

ammonia

Explanation:

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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Considering the definition of STP conditions, 4.83 × 10⁻³ moles of HF will occupy a volume of 0.108192 L at STP.

The 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.

In 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.

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Answer:

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:

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:

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:

Finally, we proceed to get the percentage yield which is calculated using the formula below:

The actual yield is the observed mass which is given as 0.88 g

The percent yield is thus:

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

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.

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;

From 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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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:

And we can rearrange it to calculate the number of moles:

Applying the values provided by the question:

Now, we can rearrange the equation of ideal gases to calculate the volume:

And, applying the values provided and the number of moles as calculated:

Therefore, the final volume of the balloon is 2.26 x 10^3 mL.

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

-----------------------------------------------------------------------------------------------------------------

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)

For H)

For O)

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

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.

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:

The 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

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