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cupoosta [38]
3 years ago
13

SHOW WORK

Chemistry
1 answer:
stealth61 [152]3 years ago
4 0
PV = nRT —> n = PV/RT

P = 2.90 atm
V = 4.80 L
R = 0.08206 L atm / mol K
T = 62.0 + 273 = 335 K (make sure you convert from celsius to kelvin)

n = (2.90 • 4.80) / (0.08206 • 335) = 0.506 moles of gas
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Two aircraft rivets, one iron and the other copper, are placed in a calorimeter that has an initial temperature of 20.°C. The da
Igoryamba

The heat will flow from Cu to Fe because Cu has higher temperature and has lower peak value .

Given ,

two aircraft rivets , one iron and the other copper , are placed in a calorimeter .

the initial temp of the calorimeter is 20 degC .

mass of iron = 30 g

initial temp of iron = 0 degC

Mass of copper = 20 g

initial temp of copper = 100 degC

Here copper has higher temp ,thus heat flow from Cu to Fe .

<h3>What is heat flow ?</h3>

Heat flow is the movement of heat from higher temperature to lower temperature .

<h3>What is calorimeter ?</h3>

It is an object used for measuring the heat flow , heat capacity of a chemical reaction as well as the physical changes .

Learn more about calorimeter here :

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8 0
1 year ago
PLEASE HELP I'M SO DOOMED
BartSMP [9]

Answer:

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Explanation:

31 / 2.8 = 11.0714286 L per mole of helium

3.5 / 4 = 0.875 moles

2.8 + 0.875 = 3.675 moles

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3 0
3 years ago
a piece of wood sinks in ethanol but floats in gasoline. what can you determine about the three different densities?
lakkis [162]
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4 0
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Read 2 more answers
Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
Anastaziya [24]

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

5 0
4 years ago
-60 points-
Firdavs [7]

Answer:

1.23e+27

Explanation:

6 0
3 years ago
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