PLEASE HELP ME!!!
A)Nuclear fusion releases protons and neutrons, so the total number of protons
and neutrons in a star changes throughout its life.
B)Nuclear fusion conserves protons and neutrons, so the total number of protons
and neutrons in a star changes throughout its life.
C)Nuclear fusion releases protons and neutrons, so the total number of protons
and neutrons in a star remains the same throughout its life.
D)Nuclear fusion conserves protons and neutrons, so the total number of protons
and neutrons in a star remains the same throughout its life.
Answers
A)Nuclear fusion releases protons and neutrons, so the total number of protons and neutrons in a star changes throughout its life is the true statement.
What are some uses for nuclear fusion?A suggested method of producing energy would use heat from nuclear fusion processes to produce electricity. A heavier atomic nucleus is created by the fusion of two lighter ones, which also produces energy. Fusion reactors are devices created to use this energy.
Nuclear fusion is less risky than nuclear fission because the fuel rods produced by nuclear fission include dangerous radioactive waste that may be used in weapons and must be maintained carefully for thousands of years.
Powering the Sun and other stars are nuclear fusion processes. Two light nuclei combine to produce one heavy nucleus during a fusion process. The resultant single nucleus's total mass is less than the mass, which causes the process to release energy.
learn more about Nuclear fusion refer
https://brainly.com/question/17870368
#SPJ13
Related Questions
Which two elements would have the same number of energy levels
Answers
The elements with the same number of energy levels are the ones that belong to the same period of the periodic table.
Copper and Zinc both belong to the 4th period of the periodic table.
Nickel belongs to the 4th period and palladium to the 5th period.
Lithium belongs to the 2nd period and Magnesium to the 3rd period.
It means that the correct answer is Copper and Zinc, they belong to the same period,
How many moles are in 53.99 mg chromium?
Answers
17 was found from determination in a mass spectrometer that an element X has three Isotopes whose mass are & 19.19,20.99 and 21.99 respectively The abundance of these I sotopes are 90.92% 0.25% $8.83% respectively Calculate the relative atomic mass.
Answers
The relative atomic mass of the given element is 40.372 amu.
What is relative atomic mass?The relative atomic mass of an element is considered as the sum of the isotopes masses each multiplied by the percentage which is found in nature.
The formula which is used to calculate the relative atomic mass is
Relative atomic mass = sum of all atomic masses of isotopes × fractional abundance
Given,
Mass of isotopes 1 = 19.19 amu
Mass of isotopes 2 = 20.99 amu
Mass of isotopes 3 = 21.99 amu
Fractional abundance of isotope 1 = 0.9092
Fractional abundance of isotope 2 = 0.0025
Fractional abundance of isotope 3 = 0.0883
By substituting all the values, we get
[( 19.19 × 0.9092) + (20.99 × 0.0025) + (21.99 × 0.883)]
= 17.447 + 0.052 + 22.873
= 40.372 amu.
Thus, we concluded that the relative atomic mass of the given element is 40.372 amu.
learn more about relative atomic mass:
https://brainly.com/question/13055840
#SPJ10
Can someone help me to answer this?As you reflect on how to interpret a balanced chemical reaction.Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 2002, Philippines has banned imports of ammonium nitrate that used in bombs that killed 12 people in Mindanao area.The explosion resulted from this reaction:2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)Construct a table showing how to interpret the information in the equation in terms of:1. individual molecules and ions.2. moles of reactants and products.3. grams of reactants and products given 2 mol of ammonium nitrate.4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
Answers
(1)
The molecule NH4NO3 is ammonium nitrate, which makes a redox reaction.
It produces N2, O2, and H2O, which are nitrogen, oxygen, and water.
The ions are 2NH4+ and 2NO3-.
2 molecules of N2, 4 molecules of H2O, and 1 molecule of O2.
(2)
There are 2 moles of NH4NO3, 2 moles of N2, 4 moles of H2O, and 1 mole of O2.
(3)
The molar mass of NH4NO3 is 80.043 grams per mole, but there are 2 moles of it, so there are 160.09 grams of NH4NO3.
There are 56.03 grams of N2 because there are 2 moles of it. (1 mole N2 = 28.0134 g/mol).
There are 72.06 grams of H2O because there are 4 moles of it (1 mole H2O = 18.02 g/mol).
There are 31.998 grams of O2 because there's just 1 mole of it.
(4)
The formula units of NH4NO3 is 1.204x10^24, which is equivalent to 2 moles.
Nitrogen has the same formula units because there are 2 moles of it, so it's 1.204x10^24.
Water has 2.409x10^24 because there are moles of it.
Oxygen has 6.022x10^23 because there's just 1 mole, inis Avogadro's Number.
What functional group is found in amino acids?A) aminesB) alkanesB) cyclic hydrocarbon ringsC) alcohols
Answers
Answer:
[tex]A)\text{ Amines}[/tex]Explanation:
Here, we want to get the functional group is found in amino acids
In the amino acids, we can get amines, carboxylic acid groups, and the carbon chain
Looking at the options, we can see that the amine group is the right option here as the other two are not available
What is the density of hydrogen sulfide (H2S) at 0.2 atm and 311 K?Answer in units of g/L
Answers
Answer
Density = 0.267 g/L
Explanation
Given:
Pressure of H2S = 0.2 atm
Temperature = 311 K
We know:
The molar mass of H2S = 34,1 g/mol
R constant = 0.08206 L.atm/K.mol
Solution:
From the ideal gas law:
PV = nRT
We know that:
density = m/V
n = m/M
Therefore we can use the following equation to solve for density of H2S
[tex]\begin{gathered} density\text{ = }\frac{PM}{RT} \\ density\text{ = }\frac{(0.2\text{ atm x 34,1 g/mol\rparen}}{(0.08206\text{ }L.atm/K.mol\text{ x 311 K\rparen}} \\ \\ density\text{ = 0.267 g/L} \end{gathered}[/tex]The quantatum mechanical model of an atom uses atomic orbitals to describe what
Answers
17. Which of the following represents a formula for a chemical compound?A. CB. KOHC. O
Answers
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)
8. Based on the Law of Conservation of Matter: At the start of the reaction 20g of
one material and some amount of another material were reacted and produced
30g of solid and 70g of a gas. What is the other amount of reactant used?
a. 20
b. 60g
c. 80g
d. 100g
Answers
Answer:
80g
Explanation:
The law of conservation of matter states that matter cannot be created nor destroyed. Basically, whatever mass you have at the beginning of a reaction, should be the same as at the end of the reaction on the product side.
Since here it lets you know that a total of 100g (30g + 70g) were produced on the product side, that means that we started off with 100g in our reactant side.
We have given that one material is 20g on our reactant side but we need the mass of the other. To find the mass of the other material, simply subtract 20g from the total mass created on the product side.
100g - 20g = 80g
The 80g would be the missing amount from the reactant side that isn't stated.
What a balanced chemical equation for the single displacement reaction you observed in Experiment 3. Include physical states.
Answers
Methane(CH4) gas and oxygen (O2) gas react to form carbon dioxide (CO2) gas and water vapor(H2O). Suppose you have 5.0 mol of CH4 and 1.0 mol of O2 in a reactor.
What would be the limiting reactant? Enter its chemical formula below
Answers
The limiting reactant, given that 5.0 moles of CH₄ and 1.0 mole of O₂ are in the reactor is O₂
How do I determine the limiting reactantWe'll begin by obtainig the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
The limiting reactant for the reaciont can be obtained as illustrated below:
From the balanced equation above,
1 mole of CH₄ reacted with 2 moles of O₂
Therefore,
5 moles of CH₄ will react with = 5 × 2 = 10 moles of O₂
From the above illustration, we can see that a higher amount of O₂ is needed to react completely with 5 moles of CH₄.
Thus, we can conclude that O₂ is the limiting reactant.
Learn more about stoichiometry:
https://brainly.com/question/11587316
#SPJ1
Ascorbic acid (vitamin C) is important in many metabolic reactions in the body, including the synthesis of collagen and prevention of scurvy. Given that themass percent composition of ascorbic acid is 40.9% C, 4.58% H, and 54.5% O, determine the empirical formula of ascorbic acid. Show all your work
Answers
Empirical formula:
Step 1
Information already provided
The mass percent composition:
40.9 % C
4.58 % H
54.5 % O
Information needed: from the periodic table
For C) 1 mol = 12.01 g
For H) 1 mol = 1.008 g
For O) 1 mol = 15.99 g
---------------------------
Step 2
A sample of 100 g is assumed, so:
40.9 % C => 40.9 g C
4.58 % H => 4.58 g H
54.5 % O => 54.5 g O
--------------------------
Step 3
Convert mass into moles:
40.9 g C x (1 mol/12.01 g) = 3.40 moles C
4.58 g H x (1 mol/1.008 g) = 4.54 moles H
54.5 g O x (1 mol/15.99 g) = 3.40 moles O
------------------------
Step 4
All moles calculated in step 3 need to be divided by the smallest one.
3.40 moles C/3.40 moles = 1
4.54 moles H/3.40 moles = 1.33
3.40 moles O/3.40 moles = 1
-----------------------
Step 5
Integer numbers are needed, so let's multiply by 3 all of them in step 4
Therefore,
For C) 3
For H) 3.99 = 4 approx.
For O) 3
All these numbers calculated will be the subindexes in ascorbic acid
Answer:
Empirical formula: C3H4O3
A reaction experimentally yields 15.68 g of a product. What is the percent yield if the theoretical yield is 18.81 g?
Answers
Answer
The percent yield = 83.36%
Explanation
Given:
Experimental yield = actual yield = 15.68 g
Theoretical yield = 18.81 g
What to find:
The percent yield for the reaction.
Step-by-step solution:
The percent yield for the reaction can be calculated using the formula below:
[tex]\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{15.68\text{ }g}{18.81\text{ }g}\times100\% \\ \\ Percent\text{ }yield=83.36\% \end{gathered}[/tex]Hence, the percent yield for the reaction is 83.36%
What is the largest possible electronegativity difference for a bond to be covalent?A.0.5B.1.7C.0.0D.1.0
Answers
Answer
B. 1.7
Explanation
As a rule, an electronegativity difference of 2 or more on the Pauling scale between atoms leads to the formation of an ionic bond. A difference of less than 2 between atoms leads to covalent bond formation.
Therefore, the largest possible electronegativity difference for a bond to be covalent is 1.7
A student finds a piece of metal and finds its mass to be 750 g. Through water displacement thestudent determines the volume to be 65.8 cm". Which metal does the student have?
Answers
Metals normally have different densities, so we can try to determine which metal is this by its density.
Assuming it is pure, the density of the metal is its mass divided by its volume:
[tex]\begin{gathered} \rho=\frac{m}{V} \\ \rho=\frac{750g}{65.8cm^3} \\ \rho\approx11.4g/cm^3 \end{gathered}[/tex]Now, we need to look for some table with densities of various metals and see which one has the density we found, 11.4 g/cm³.
Since we don't have one, we can look for one. In it, we can see that the only metal with this density is lead.
So, the metal in the question should be lead.
If you wanted to dilute the 3M NaOH solution to 500mL of 1M NaOH solution, how much L of the 3M NaOH solution would you need?
Answers
0.167L
Explanation:In order to know how much L of the 3M NaOH solution would you need, we will simply set up an equal proportion solution expressed as;
[tex]C_1V_1=C_2V_2[/tex]C1 and C2 are the concentration of the solutions
V1 and V2 are the volumes of the solutions
Given the following parameters;
C1 = 1M
V1 = 500mL
C2 = 3M
V2 = ?
Substitute the given parameters into the formula above to get the required litre of solution needed.
[tex]\begin{gathered} 1\times500=3\times V_2_{} \\ 500=3V_2 \\ \text{Swap} \\ 3V_2=500 \end{gathered}[/tex]Divide both sides by 3:
[tex]\begin{gathered} \frac{3V_2}{3}=\frac{500}{3}_{} \\ V_2=166.67mL \end{gathered}[/tex]Converting to litres
Since 1mL = 0.001L
166.67mL = x
Cross multiply
[tex]\begin{gathered} 1\times x=166.67\times0.001 \\ x=0.167L \end{gathered}[/tex]Hence the amount of L of the 3M NaOH solution would you need is 0.167L
Classify CH3CH2NH2 as astrong base or a weak base.Strong BaseWeak Base
Answers
Answer:
CH3CH2NH2 is a weak base.
Explanation:
CH3CH2NH2 is a weak base since its Kb is small, and thus it partially dissociates.
A solution with a total volume of 1000.0 mL contains 37.1 g Mg(NO3)2. If you remove 20.0 mL of this solution and then dilute this 20.0 mL sample with water until the new volume equals 500.0 mL, what is the concentration of Mg+2 ion in the 500.0 mL of solution? What is the concentration of nitrate ion?
Answers
1. The concentation of the magnesium ion, Mg²⁺ in the solution is 0.01 M
2. The concentation of the nitrate ion, NO₃⁻ in the solution is 0.02 M
We'll begin by obtaining the concentration of the stock solution. This can be obtained as follow:
Mass of Mg(NO₃)₂ = 37.1 gMolar mass of Mg(NO₃)₂ = 148 g/moleMole of Mg(NO₃)₂ = 37.1 / 148 = 0.25 moleVolume = 1000 mL = 1000 / 1000 = 1 LConcentration =?Concentration = mole / volume
Concentration = 0.25 / 1
Concentration = 0.25 M
Next, we shall determine the concentration of the diluted solution
Volume of stock solution (V₁) = 20 mLConcentration of stock solution (C₁) = 0.25 MVolume of diluted solution (V₂) = 500 mL Concentration of diluted solution (C₂) =?C₁V₁ = C₂V₂
0.25 × 20 = M₂ × 500
5 = M₂ × 500
Divide both side by 500
C₂ = 5 / 500
C₂ = 0.01 M
1. How to determine the concentration of magnesium ion, Mg²⁺
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 1 mole of Mg²⁺
Therefore,
0.01 M Mg(NO₃)₂ will also contains 0.01 M Mg²⁺
Thus, the concentration of Mg²⁺ is 0.01 M
2. How to determine the concentration of nitrate ion, NO₃⁻
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 2 mole of NO₃⁻
Therefore,
0.01 M Mg(NO₃)₂ will contain = 0.01 × 2 = 0.02 M NO₃⁻
Thus, the concentration of NO₃⁻ is 0.02 M
Learn more about dilution and concentration:
https://brainly.com/question/15022582
https://brainly.com/question/13565111
#SPJ1
Describe global influences on local
weather.
Answers
Substance Density (grams/cm3)Chloroform - 1.5Ebony wood - 1.2Mahogany wood - 0.85Oil - 0.9Water - 1.025.Since volume = mass/density, a 1,700 gram beam of mahogany wood has a volume of...Volume = Mass / DensitySelect one:a. 500 cm3b. 1,445 cm3c. 1,785 cm3d. 2,000 cm3
Answers
As the question gave us the formula in which we have to use to calculate the volume of this type of wood:
V = m/d
We have:
m = 1700 grams
d = 0.85
Now we add these values into the formula:
V = 1700/0.85
V = 2000 cm3, letter D
The following lists consists of ionic compounds EXCEPT
barium hydroxide, zinc carbonate, ammonium sulfate
calcium chloride, carbon disulfide, magnesium nitrate
sodium sulfate, copper(II) oxide, potassium nitride
aluminium sulfide, sodium sulfite, calcium fluoride
Answers
The following lists consists of ionic compounds except carbon disulfide (CS₂).
Barium hydroxide , Ba(OH)₂ is an ionic compound.
zinc carbonate, ZnCO₃ is an ionic compound.
ammonium sulfate , (NH₄)₂SO₄ is an ionic compound.
calcium chloride, CaCl₂ is an ionic compound.
carbon disulfide, CS₂ is not an ionic compound. In carbon disulfide both the elements are non metallic elements. The bond formed between atoms are by sharing of electron known as covalent bond due to very little difference in electronegativity.
magnesium nitrate, Mg(NO₃)₂ is an ionic compound.
sodium sulfate, Na₂SO₄ is an ionic compound.
copper(II) oxide, CuO is an ionic compound.
potassium nitride KNO₃ is an ionic compound.
aluminium sulfide, Al₂S₃ is an ionic compound.
sodium sulfite, Na₂S is an ionic compound.
calcium fluoride, CaF₂ is an ionic compound.
Thus, The following lists consists of ionic compounds except carbon disulfide (CS₂).
To learn more about ionic compound here
https://brainly.com/question/9167977
#SPJ1
How many grams of H2 are required to completely convert 80g of Fe2O3?
Answers
Answer
3.0 grams H₂ is required.Explanation
Given:
Mass of Fe2O3 = 80 g
Equation:
What to find:
The grams of H2 required to completely convert 80g of Fe2O3.
Step-by-step solution:
From the equation of reaction;
3 moles of H2 completely react with 1 mole of Fe2O3
Note: Molar mass of H2 is 2.016 grams per mole and Molar mass of Fe2O3 is 159.69 g/mol
This implies; (3 x 2.016 g) = 6.048 grams H2 completely react with 159.69 grams Fe2O3.
Therefore, x grams H2 will completely convert 80 grams Fe2O3.
Cross multiply and divide both sides by 159.69 grams Fe2O3.
x grams H2 is now equal to
[tex]x=\frac{80\text{ }g\text{ }Fe_2O_3}{159.69\text{ }g\text{ }Fe_2O_3}\times6.048\text{ }g\text{ }H_2=3.0298\approx3.0\text{ }grams\text{ }H_2[/tex]Therefore the grams of H2 required to completely convert 80g of Fe2O3 is 3.0 grams
Starting with a gas of N2 in a balloon of temperature 148.5°C and volume 241.8mL, what is its final volume if you cool it to -96.4°C?
Answers
Answer
101.3 mL
Explanation
Given:
The initial temperature, T₁ = 148.5 °C = (148.5°C + 273) = 421.5 K
The initial volume, V₁ = 241.8 mL
Final temperature, T₂ = -96.4 °C = (-96.4°C + 273) = 176.6 K
What to find:
The final volume of the gas.
Step-by-step solution:
The final volume, V₂ of the gas can be calculated using Charle's law formula.
[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \\ \Rightarrow V_2=\frac{V_1\times T_2}{T_1}=\frac{241.8mL\times176.6K}{421.5K}=\frac{42701.88\text{ }mL}{421.5} \\ \\ V_2=101.3\text{ }mL \end{gathered}[/tex]The final volume of the gas is 101.3 mL
How to find oxidation number of ReO4^-?
Answers
The oxidation number of O is usually -2 and in this case it is.
Our ion is ReO₄⁻. It has a total charge of -1. The charge of the four atoms of O is -8. With this information we can write an equation. X will represent the oxidation state of Re.
Total charge = Oxidation state of Re + 4 * Oxidation state of O
- 1 = x + 4 * (-2)
- 1 = x - 8
-1 + 8 = x
x = +7
So the oxidation state of Re is +7.
Sugar forms when carbon, oxygen, and hydrogen combine in a specific ratio. From what you know about elements and the periodic table, what is true about the bonding in sugar?
Answers
Answer
B. The bonds in the compound are covalent
Explanation
The carbon, oxygen, and hydrogen that combine in a specific ratio to form the sugar are nonmetals. From what was learned about elements and the periodic table, nonmetals form covalent bonds among themselves to form compounds.
So what is true about the bonding in sugar is:
B. The bonds in the compound are covalent.
I’m not sure and I’m kind of confused can anyone help?
Answers
We will reconstruct the model in the following manner :
From the above diagram we can see that :
• number of Carbon atom = 3
• number of hydrogen atom = 8
• rewrite this in an alphabetical order, you get :
[tex]\begin{gathered} C_3H_8\text{ } \\ \Rightarrow Propane\text{ } \end{gathered}[/tex]the molecule has a chemical formula = C3H8How many moles of chromium metal, Cr, are in a 260 gram piece of chromium?
Answers
In order to answer this question we will use the molar mass of Chromium, which is 52 g/mol, that means that in every 1 mol of Cr, we will have 52 grams of it:
52 g = 1 mol
260 g = x moles
52x = 260
x = 5 moles of Chromium
How many molecules of ethane gas, C2H6 are in 15 grams of the compound?
Answers
3.01×10²³molecules.
Explanations:
The formula for the number of molecules of a compound given the number of moles is expressed as:
[tex]nu\text{mber of molecules=moles}\times6.02\times10^{23}[/tex]Get the moles of ethane gas using the formula:
[tex]\begin{gathered} \text{moles of ethane=}\frac{Mass\text{ of ethane}}{Molar\text{ mass of ethane}} \\ \text{Moles of ethane=}\frac{15}{2(12)+1(6)} \\ \text{Moles of ethane}=\frac{15}{30}\text{moles} \\ \text{Moles of ethane}=0.5\text{moles} \end{gathered}[/tex]Determine the required number of molecules of ethane
[tex]\begin{gathered} nu\text{mber of mol}ecules=0.5\times6.02\times10^{23} \\ nu\text{mber of mol}ecules=3.01\times10^{23} \end{gathered}[/tex]Hence the molecule of ethane gas that is in 15 grams of the compound is 3.01×10²³molecules.
Determine the percent composition of hydrogen for the following: NaHCO3
Answers
By definition, the percent composition of an atom in a compound is its mass percentage in the formula.
That is, if we have 1 mol of NaHCO₃, we have also 1 mol of H (because there is only on H for each molecule).
So, we calculate the mass of this 1 mol of NaHCO₃ and the mass of 1 mol of H and calculate the percentage.
In equations, we want the following:
[tex]C_H=\frac{m_H}{m_{NaHCO_{3}}}[/tex]Since these are ratios, we doesn't matter if we talk about 1, 2 or any number of moles, but 1 mol is easier because the molecular and atomi masses are for 1 mol.
The molecular mass of NaHCO₃ is:
[tex]\begin{gathered} M_{NaHCO_3}=M_{Na}+M_H+M_C+3\cdot M_O \\ M_{NaHCO_3}\approx(22.990+1.008+12.011+3\cdot15.999)g/mol \\ M_{NaHCO_3}\approx84.006g/mol \end{gathered}[/tex]Which means that we have approximately 84.006 grams of NaHCO₃ in 1 mol of it.
The atomic mass of H is:
[tex]M_H\approx1.008g/mol[/tex]Which means that we have approximately 1.008 grams of H in 1 mol of it.
Now, we can take the percentage of mass of H:
[tex]C_H\approx\frac{1.008g_{}}{84.006g}\cdot100\%\approx1.20\%[/tex]So, the percentage composition of H in NaHCO₃ is approximately 1.20%.
54000 mL isO 54 m3O 5400 cm3O 0.054m3O 540 m3
Answers
mL and cm³ have a 1 to 1 conversion, so we have:
[tex]54000mL=54000cm^{3}[/tex]But we don't have this option, so we will need to find another.
The other are in m³, so we can use the conversion from cm to m to get this:
[tex]\begin{gathered} 1m=100cm \\ (1m)^{3}=(100cm)^{3} \\ 1m^{3}=1000000cm^{3} \\ 1cm^{3}=\frac{1}{1000000}m^{3} \end{gathered}[/tex]So, we can apply this to what we have:
[tex]54000cm^3=54000\cdot\frac{1}{1000000}m^3=0.054m^{3}[/tex]We have an option with 0.054m³, so the correc alternative is 0.054 m³.