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Mashutka [201]
3 years ago
12

The temperature of a body of water influences

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
6 0

the temperature of the air above it

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Please help me urgent!
loris [4]
It decreases in volume, as the energy is lost.
4 0
3 years ago
Given the equation:
Citrus2011 [14]

Answer: a)  Fe is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al  

To calculate the moles :

   

\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles

\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles

According to stoichiometry :

a) 9 moles of Fe  require= 4 moles of Al_2O_3

Thus 0.15 moles of Fe will require=\frac{4}{9}\times 0.15=0.067moles  of Al

Thus Fe is the limiting reagent as it limits the formation of product and Al is the excess reagent.

b) As 9 moles of Fe give = 8 moles of Al

Thus 0.15 moles of Fe give =\frac{8}{9}\times 0.15=0.133moles  of Al

Mass of Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g

c) As 9 moles of Fe give = 3 moles of Fe_3O_4

Thus 0.15 moles of Fe give =\frac{3}{9}\times 0.15=0.05moles  of Fe_3O_4

Mass of Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g

8 0
3 years ago
A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/
Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0
3 years ago
True or false help please I’ll give you brainly
dimulka [17.4K]
The answer is true.
5 0
3 years ago
Read 2 more answers
The final answers for the following problems should be written
ladessa [460]

Answer:

2.24725 kilograms

Explanation:

To convert grams to kilograms, you divide the number of grams you have by 1000. So, 800 g = 2.225 x 1010/1000 = 2.24725 kg

8 0
2 years ago
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