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MissTica
4 years ago
9

What does the number tell us about an atom that never changes

Chemistry
1 answer:
Leokris [45]4 years ago
6 0
The number that never changes is the atomic mass (atomic number on the periodic table)
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Choose the best classification of the reactions C6H12O6 (g) + 6O2 yields 6CO2 + 6H2O
marta [7]
Combustion reaction because it produces water and carbon dioxide
6 0
3 years ago
How is synthesis and decomposition reaction related to each other???
Anit [1.1K]
They are kind of opposite processes. Chemical synthesis is execution of chemical reactions to make products. Like you take two substances, you put them together, they react and you have new a substance. For example. 2Na+{ Cl }_{ 2 }\rightarrow 2NaCl here Sodium and Chlorine react to make a new substance which is Sodium Chloride.

Chemical decomposition is process of seperating a substance to different substances. There is a substance by a reaction this substance becomes two substances, Like : XY -> X+Y For example : 2{ H }_{ 2 }O(l)\quad \rightarrow 2{ H }_{ 2 }+{ O }_{ 2 }
8 0
3 years ago
Read 2 more answers
Which of the following reactions will have the largest equilibrium constant (K) at 298 K? Which of the following reactions will
hram777 [196]

Answer:

<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>

Explanation:

If we know the ΔG° of a chemical reaction it is possible to calculate the equilibrium constant (k) of this procedure with the next equation:

ln Keq = -ΔG° / RT (1)

Where: Keq is equilibrium contant, ΔG° is standard state free energy change, R is gas constant and T is temperature.

Watching (1), it is possible to know that the large negative ΔG° the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of equation it is necessary that ΔG° be negative.

Knowing this, is the oxidation of Hg the reaction that has the largely negative ΔG°. So, this reaction will have the largest equilibrium constant.

<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>

CaCO₃(s) → CaO(s) + CO₂(g) ΔG° =+131.1 kJ

3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) ΔG° = -28.0 kJ

I hope it helps!

8 0
3 years ago
2. Iron reacts with oxygen gas according to the following equation:
erastovalidia [21]

Answer:

2. a) 2.67 mol.

  b) 1.33 mol.

3. 4.35 g.

4. 8.67 g.

5. a) 143.86 L.

   b) 251.75 L.

Explanation:

<em>2. Iron reacts with oxygen gas according to the following equation: </em>

<em>4Fe + 3O₂ → 2Fe₂O₃ , If 2 moles of oxygen gas is used in the reaction, </em>

  • Fe reacts with O₂ according to the balanced equation:

<em>4Fe + 3O₂ → 2Fe₂O₃,</em>

It is clear that 4 mole of Fe react with 3 mole of O₂ to  produce 2 moles of Fe₂O₃.

a) how many moles of iron, Fe, will be required?

<u><em>using cross multiplication: </em></u>

3 mol of O₂ require  → 4 mol of Fe, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe.

∴ The no. of moles of of Fe are required = (2 mol)(4mol)/(3 mol) = 2.67 mol.

(b) how many moles of iron(III) oxide, Fe₂O₃, will be produced?

<u><em>using cross multiplication: </em></u>

3 mol of O₂ produce → 2 mol of Fe₂O₃, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe₂O₃.

∴ The no. of moles of of Fe₂O₃ are produced = (2 mol)(2 mol)/(3 mol) = 1.33 mol.

<em>3. Potassium sulfate can be prepared by the reaction between dilute sulfuric acid  and potassium carbonate. </em>

<em>H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O, </em>

<em>Calculate the mass of potassium sulfate that can be prepared from 3.45 g of  potassium carbonate.</em>

  • H₂SO₄ reacts with K₂CO₃ according to the balanced equation:

H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O<em>,</em>

It is clear that 1 mole of H₂SO₄ reacts with 1 mole of K₂CO₃ to  produce 1 mole of K₂SO₄, 1 mole of CO₂ and 1 mole of H₂O.

Firstly, we need to calculate the no. of moles of 3.45 g of K₂CO₃:

no. of moles of K₂CO₃ = mass/molar mass = (3.45 g)/(138.205 g/mol) = 0.025 mol.

<u><em>using cross multiplication: </em></u>

1 mol of K₂CO₃ produce → 1 mol of K₂SO₄, from the stichiometry.

∴ 0.025 mol of K₂CO₃ produce → <em>0.025 mol of K₂SO₄.</em>

∴ The mass of K₂SO₄ are produced = (no. of moles of K₂SO₄ produced)(molar mass of K₂SO₄) = (0.025 mol)(174.259 g/mol) = 4.35 g.

4. The reaction between zinc and aqueous chromium(III) nitrate can be represented  by the following equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr

If 16.25 g of zinc is used to react with chromium(III) nitrate, calculate the mass of  chromium that will be produced.

  • Zn reacts with Cr(NO₃)₃ according to the balanced equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr <em>,</em>

It is clear that 3 mole of Zn reacts with 2 mole of Cr(NO₃)₃ to  produce 3 mole of Zn(NO₃)₂  and 2 mole of Cr.

Firstly, we need to calculate the no. of moles of 16.25 g of Zn:

no. of moles of Zn = mass/atomic mass = (16.25 g)/(65.38 g/mol) = 0.25 mol.

<u><em>using cross multiplication: </em></u>

3 mol of Zn produce → 2 mol of Cr, from the stichiometry.

∴ 0.25 mol of Zn produce → <em>??? mol of Cr.</em>

∴ The no. of moles of Cr are produced = (2 mol)(0.25 mol)/(3 mol) = 0.167 mol.

∴ The mass of Cr are produced = (no. of moles of Cr produced)(atomic mass of Cr) = (0.167 mol)(51.9961 g/mol) = 8.67 g.

<em>5. Ethane, C₂H₆, burns in oxygen gas according to the following equation: </em>

<em>2 C₂H₆ + 7 O₂ → 4 CO₂ + 6H₂O, </em>

<em>If 72 dm³ of ethane gas is completely burnt in oxygen, calculate </em>

(a) the volume of carbon dioxide, measured at room temperature and

pressure produced.

Firstly, we can calculate the no. of moles of 72 dm³ ethane at room temperature and pressure using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm).

V is the volume of the gas in L (V = 72.0 dm³ = 72.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 298.0 K, room temperature).

∴ n of ethane = PV/RT = (1.0 atm)(72.0 L)/(0.0821 L.atm/mol.K)(298.0 K) = 2.94 mol.

So, we can calculate the no. of moles of CO₂:

<u><em>using cross multiplication: </em></u>

2 mol of C₂H₆ produce → 4 mol of CO₂, from the stichiometry.

∴ 2.94 mol of C₂H₆  produce → <em>??? mol of </em>CO₂<em>.</em>

∴ The no. of moles of CO₂ are produced = (2.94 mol)(4.0 mol)/(2 mol) = 5.88 mol.

∴ The volume of moles of CO₂ are produced = nRT/P = (5.88 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 143.86 L.

(b) the volume of oxygen, measured at room temperature and pressure

required​.

<u><em>using cross multiplication: </em></u>

2 mol of C₂H₆ require → 7 mol of O₂, from the stichiometry.

∴ 2.94 mol of C₂H₆  require → <em>??? mol of </em>O₂<em>.</em>

∴ The no. of moles of O₂ are required = (2.94 mol)(7.0 mol)/(2 mol) = 10.29 mol.

∴ The volume of moles of O₂ are produced = nRT/P = (10.29 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 251.75 L.

8 0
3 years ago
HELP ME FAST PLEASE
Katena32 [7]

Answer:

B

Explanation:

when two plates come together and one goes underneath, it forms a ridge.

3 0
3 years ago
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