Can someone help me to answer these Chemistry questions?

On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical

zhuklara [117]

Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

5 0

2 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

Which of the following has the largest radius? Answer:K

mixas84 [53]

Answer:

the answer is k? lol thanks

4 0

3 years ago

How many atoms are found in 3.45g of CO2?

USPshnik [31]

Answer: The number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol$

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains $6.022\time 10^{23}$ number of molecules

So, 0.0784 moles of carbon dioxide gas will contain $1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}$ number of carbon atoms and $2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}$ number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

3 0

3 years ago

What is the charge of a chlorine ion that has gained 1 electron?

nikdorinn [45]

schoOOOL SUcKS sO BADDDDDD

7 0

3 years ago

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