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Andrews [41]
3 years ago
12

What type of reaction is "Separating Water"?

Chemistry
1 answer:
Lostsunrise [7]3 years ago
3 0

Answer:

Decomposition

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Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0
3 years ago
What does one mole of oxygen and one mole of carbon dioxide have in common?
arsen [322]

Answer:

Equal number of moles

Explanation:

Thats life

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3 years ago
Where would a primary producer be located in a food chain?
vfiekz [6]

Answer:

The primary producer would be at the bottom of the food chain.

Explanation:

4 0
4 years ago
Read 2 more answers
Idk y my teacher thinks Ik what this is so I need help asap
lara [203]
2 H₂ + O₂ −−−> 2 H₂O
6 0
4 years ago
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
3 years ago
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