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julsineya [31]
3 years ago
13

Determine the molarity for the fixer Sodium carbonate decahydrate (I dissolved 127.62 grams of sodium carbonate for every 4 lite

rs of water)
Chemistry
1 answer:
grin007 [14]3 years ago
4 0

Answer:

0.112 M.

Explanation:

  • Molarity is the no. of moles of solute in a 1.0 L of the solution.

M = n/V.

<em>M = (mass/molar mass)solute x (1000/V of the solution).</em>

mass = 127.62 g.

molar mass = 286.138 g/mol.

V of the solution = 4.0 L = 4000.0 mL.

<em>∴ M = (mass/molar mass)solute x (1000/V of the solution)</em> = (127.62 g / 286.138 g/mol) x (1000 / 4000.0 mL) = <em>0.1115 M ≅ 0.112 M.</em>

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An ate or ite at the end of a compound name usually indicates that the compound contains___________.
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Explanation:

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As in carbonate (CO3^2-)

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Why did the plant that could not photosynthesize stop growing?
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3 years ago
In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is
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Answer:

the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

Explanation:

Given that:

mass of diphenylacetylene (C_{14}H_{10}) = 0.5297 g

Molar Mass of diphenylacetylene (C_{14}H_{10}) = 178.21 g/mol

Then number of moles of diphenylacetylene (C_{14}H_{10})  = \frac{mass}{molar \ mass}

= \frac{0.5297  \ g }{178.24 \  g/mol}

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = Heat absorbed by H_2O + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene (C_{14}H_{10}) will be = \frac{20598.87 \ J}{0.002972 \ mol}

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Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

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