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Vikentia [17]
3 years ago
9

For the reaction HNO3 + Mg(OH)2→ Mg(NO3)2 + H2O, how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid,

HNO3? Balance the equation
Chemistry
2 answers:
ollegr [7]3 years ago
6 0

Answer:

tha mass of magnesium nitrate is 592g

Explanation:

from a balanced chemical equation

2HNO3 + Mg(OH)2→ Mg(NO3)2 + 2H2O

2 mol of nitric acid is equivalent to 1 mol of magnesium nitrate. then 8 mol of nitric acid will be equivalent to 4 mol of magnesium nitrate.

oee [108]3 years ago
5 0

Answer:

2HNO3 + Mg(OH)2 --> 2H2O + Mg(NO3)2

593.2 grams Mg(NO3)2

Explanation:

First, balance the equation

The first thing I notice is that there are 2 nitrates on the reactant side so I would put a 2 in front of HNO3. Now the problem is that there are 4 hydrogens on the reactants side, so add a 2 to H2O.

Next is stoichiometry

We are looking for Mg(NO3)2 so use the balanced equation to create a mole ratio and then convert to grams.

8 mol HNO3 * 1 mol Mg(NO3)2/2 mol HNO3 * 148.3 g Mg(NO3)2/1 mol Mg(NO3)2 = 593.2 grams Mg(NO3)2

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1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does th
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1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
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3 years ago
A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to
nasty-shy [4]

The volume at 100 mmHg : 0.656 L

<h3>Further explanation</h3>

Boyle's Law  

<em>At a constant temperature, the gas volume is inversely proportional to the pressure applied  </em>

\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

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P₁=2.5 kPa=18,7515 mmHg

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4 0
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A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
kaheart [24]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

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C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

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