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Zielflug [23.3K]
3 years ago
6

Write the fraction of the mass of kcl produced from 1 g of k2c03​

Chemistry
1 answer:
Minchanka [31]3 years ago
6 0

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

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Through hypothesis and experiments
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How many grams of sodium ions are in na2so4?
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142.04

Explanation:

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
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Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
HELP!!!!!
faltersainse [42]

Answer:

5.17.

Explanation:

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

[OH⁻] = 1.5  x 10⁻⁹ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5  x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17. </em>

5 0
3 years ago
Read 2 more answers
How many milligrams are in 2.89 x 1024 CO2 molecules?
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