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FinnZ [79.3K]
3 years ago
10

Based on the balanced chemical equation for the preparation of malachite, what is the composition of the bubbles formed when sod

ium carbonate is added to the solution of copper sulfate?
Chemistry
1 answer:
spin [16.1K]3 years ago
3 0

Answer:

Carbon Dioxide = CO2

Explanation:

The synthesis of Malachite is seen in the chemical formula:

CuSO 4 . 5H2O(aq) + 2NaCO3(aq) --> CuCO 3 Cu(OH) 2 (s) + 2Na 2 SO 4 (aq) + CO 2 (g) + 9H 2 O(l)

The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase  (g).

Upon examining the chemical equation, only CO2 is in the gaseous state and hence the only one that can be formed as bubbles,

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
2 years ago
A block of iron has a mass of 826 g. What is the volume of the block of iron whose density at 25°C is 7.9
LuckyWell [14K]

Answer:

105

Explanation:

Density = mass/volume

therefore volume = mass/density

826/7.9=105 (To three significant figure)

8 0
2 years ago
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
What is the democritus findings?
Volgvan

nvm

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4 0
2 years ago
A sample of gas at 1.10 atm has a volume of 326 mL. What is the new volume if the pressure is changed to 1.90 atm?
Brut [27]

Answer

For this we use ideal gas equation which is:

P1V1 = P2V2

P1 = 1.10 atm

V1 = 326 ml

P2 = 1.90

V2 = ?

By rearranging the ideal gas equation:

V2 =  P1V1 ÷ P2

V2 = 1.10 × 326 ÷1.90

V2 = 358.6 ÷ 1.90

V2 = 188.7 ml

8 0
3 years ago
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