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2 years ago

P4(s) + O2(g) → __ P4O10(s) Now we will balance O. How many O2 molecules are needed to form one P4O10 molecule?

Chemistry

1 answer:

Mnenie [13.5K]2 years ago

4 0

Answer:

Five molecules of oxygen.

Explanation:

Hello,

In this case, considering the given chemical reaction, we must write a five before the oxygen in order to equal the number of oxygen atoms at both the right and left side (ten) so phosphorous remain the same (4):

$P_4(s) + 5 O_2(g)\rightarrow P_4O_{10}(s)$

It means that five molecules of oxygen (O₂) are needed to form one molecule of tetraphosphorous decaoxide (P₄O₁₀).

Best regards.

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The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a

shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone = 38 %

(S) - carvone = 62%

7 0

1 year ago

Read 2 more answers

When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?

Afina-wow [57]

Answer:

$Y=75.6\%$

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

$m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6$

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

$Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%$

Best regards.

3 0

1 year ago

Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo

sleet_krkn [62]

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is favorable when ΔG < 0 (exergonic). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

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1 year ago

Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are

kolbaska11 [484]

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution = 50.0 mL = 0.050 L

Moles of copper sulfate = n

$1.00M=\frac{n}{0.050 L}$

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

$1\times 0.050 mol=0.050 mol$ of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

$CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)$

d) $CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)$ ..[1]

$KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)$ ..[2]

$Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)$

Common ion both sides are removed. The net ionic equation is given as:

$Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)$

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

$m=d\time V=1 g/ml\times 100 mL = 100 g$

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

$Q=mc\Delta T$

$Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ$

Enthalpy of the reaction = ΔH = $\frac{Q}{\text{Moles of copper}}$

ΔH = $\frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol$

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

7 0

1 year ago

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allochka39001 [22]

Explanation:

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2 years ago

__ P4(s) + __ O2(g) → __ P4O10(s) Now we will balance O. How many O2 molecules are needed to form one P4O10 molecule?

P4(s) + O2(g) → __ P4O10(s) Now we will balance O. How many O2 molecules are needed to form one P4O10 molecule?