A readily corroded silvery solid that fizzes with water

0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A

elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

$Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J$

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = $\frac{0.9975 g}{56 g/mol}=0.01781 mol$

The enthlapy of solution :

$\frac{Q'}{moles}$

$=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol$

The enthlapy of solution is -55.23 kJ/mol.

8 0

2 years ago

2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta

Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

$\eta = \frac{m}{M}$

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

$\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles$

$\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles$

$\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles$

$\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles$

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

$\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles$

We have that between MnO₂ and KOH, the limiting reactant is KOH.

$\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles$

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

$\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles$

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0

2 years ago

Read 2 more answers

In your body, carbon dioxide, CO2, dissolves in the blood to form carbonic acid, H2CO3 as follows:

Lubov Fominskaja [6]

Answer:

STOP USING THIS SIGHT STOP

Explanation:

not good for ua quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

a quart of stain covers 100 square ft how many quarts should you buy to stain the ramp assume you dont have to stain the bottem its lenth is 7 ft 1 side is 35ft the other 35 1/6 ft and the hight is 35 ft

3 0

2 years ago

A food web showing the flow of energy through a freshwater ecosystem is shown below. Which of the animals shown in the food web

kati45 [8]

Answer: minnows

Explanation: tiny fish

7 0

2 years ago

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Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As

Crank

Answer:

1.8 × 10⁻¹⁶ mol

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

1.2 + 3s 2s

$K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\$

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0

2 years ago