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FromTheMoon [43]
2 years ago
13

(pls help me)

Chemistry
2 answers:
kari74 [83]2 years ago
7 0
Claim 1 Sounds more convincing then claim 2
KATRIN_1 [288]2 years ago
5 0

Answer:

Claim 1

Explanation:

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Chemical energy sources are often portable and provide large amounts of energy. Fossil fuels are one kind of chemical energy sou
konstantin123 [22]

Answer:You can reduce our carbon footprints by using renewable energy,

Explanation:

4 0
2 years ago
What is the mass of 0.60 moles of Al
KIM [24]

16 g. The mass of 0.60 mol Al is 16 g.

Molar mass of Al = 26.98 g/mol

Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al

6 0
3 years ago
Read 2 more answers
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
Kryger [21]

Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

8 0
3 years ago
6-11<br>don't answer if you don't know​
Bumek [7]

Answer:

ghrtdfghtyfrhjbfyrt

Explanation:

t6n yrtnyrt tryrbhyrty ryrhtr

6 0
3 years ago
What is the pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at
lyudmila [28]

The pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at 145 kPa and 176°C is 1.195 atm.

<h3>What is ideal gas equation?</h3>

Ideal gas equation of any gas will be represented as:
PV = nRT, where

P = pressure

V = volume

n = moles

R = universal gas constant

T = temperature

First we calculate the moles of gas, when the volume of gas 875 ml at

145 kPa and 176°C as:

n = (1.431atm)(0.875L) / (0.082L.atm/K.mol)(449.15K)

n = 1.252 / 36.83 = 0.033 moles

Now we measure the pressure of 0.033 moles of gas of 1.27 L of a gas at 288°C as:

P = (0.033mol)(0.082L.atm/K.mol)(561K) / (1.27L) = 1.195 atm

Hence required pressure of gas is 1.195 atm.

To know more about ideal gas equation, visit the below link:
brainly.com/question/555495

#SPJ1

3 0
2 years ago
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