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kobusy [5.1K]
3 years ago
6

You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
4 0
1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.
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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}

= 6930979.139 J/mol

= 6930.98 kJ/mol

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

3 0
3 years ago
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Need help asap!!!!<br> will mark brainliest!!!
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Answer:

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Explanation:

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