Balance the following equation Sb2S3+O3=Sb2O3+SO2

Which of these scenarios involve a reaction that is at equilibrium

Tasya [4]

Reaction is producing more reactants than products

3 0

3 years ago

You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

Best regards!

7 0

3 years ago

0.5 gm of mixture of NH4Cl and NaCl was boiled with 25 ml of 0.95 N NaOH in a vessel till all the ammonia is expelled.The residu

GalinKa [24]

Answer:

72.66%

Explanation:

NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:

NH₄Cl + NaOH → NaCl + NH₃ + H₂O

The residual NaOH reacts with H₂SO₄ as follows:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

The equivalent-gram of H₂SO₄ are:

16mL * 0.1N * 1.06 = 1.696mEq.

As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

1.696mEq * (100mL / 10mL) = 16.96mEq.

The mEq of NaOH you add in the first are:

25mL * 0.95mEq = 23.75mEq

That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

6.79x10⁻³ moles NH₄Cl

In grams (Using molar mass NH₄Cl = 53.5g/mol):

6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =

0.3633g of NH₄Cl are in the original mixture.

% is:

0.3633g/ 0.5g * 100 = 72.66%

8 0

3 years ago

When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th

maxonik [38]

I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture.

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>

8 0

3 years ago

The pressure on 150 mL of a gas is increased from 500 mm Hg to 700 mm Hg at constant temperature. What is the new volume of the

Ivanshal [37]

Answer:

Option A. 107 mL

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 150 mL

Initial pressure (P₁) = 500 mmHg

Final pressure (P₂) = 700 mmHg

Temperature = constant

Final volume (V₂) =?

The final volume of the gas can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

500 × 150 = 700 × V₂

75000 = 700 × V₂

Divide both side by 700

V₂ = 75000 / 700

V₂ = 107 mL

Therefore, the final volume of the gas is 107 mL.

4 0

3 years ago