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marta [7]
3 years ago
9

The density of lead is 11.3

Chemistry
2 answers:
brilliants [131]3 years ago
8 0
Volume = 287 - 206 
v= 81 mL 

Mass= density x volume 
m = 11.3 x 81 
m = 915.3 g



wariber [46]3 years ago
8 0

Answer:

915 g

Explanation:

The volume of lead (V) is equal to the difference between the final level of water and the initial level of water.

V = 287 mL - 206 mL = 81.0 mL

The density of lead (ρ) is 11.3 g/mL. We can find the mass of lead (m), in grams, using the following expression.

ρ = m/V

m = ρ × V

m = 11.3 g/mL × 81.0 mL

m = 915 g

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3 years ago
In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery. Fe(s) + 2 NiO(OH)(s)
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Ans: Moles of Fe(OH)2 produced is 5.35 moles.


Given reaction:

Fe(s) + 2NiO(OH) (s) + 2H2O(l) → Fe(OH)2(s) + 2Ni(OH)2(aq)

Based on the reaction stoichiometry:

1 mole of Fe reacts with 2 moles of NiO(OH) to produce 1 mole of Fe(OH)2

It is given that there are:

5.35 moles of Fe

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4 0
4 years ago
Read 2 more answers
Can someone answer dis:
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Explanation:

5 0
3 years ago
When a 3.80 g sample of C8H18(l) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 27.3 oC. The heat
Fudgin [204]

<u>Answer:</u> The enthalpy of the reaction is -5112.5 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 6.18 kJ/°C

\Delta T = change in temperature = 27.3°C

Putting values in above equation, we get:

q=6.18kJ/^oC\times 27.3^oC=168.714kJ

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of octane = 3.80 g

Molar mass of octane = 114 g/mol

Putting values in above equation, we get:

\text{Moles of octane}=\frac{3.80g}{114g/mol}=0.033mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta E=\frac{q}{n}

where,

q = amount of heat released = -168.714 kJ

n = number of moles = 0.033 moles

\Delta E = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta E=\frac{-168.714kJ}{0.033mol}=-5112.5kJ/mol

Hence, the enthalpy of the reaction is -5112.5 kJ/mol

4 0
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What trend does the first ionization energy follow, going down the periodic table?
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