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Sidana [21]
3 years ago
10

Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3 (taking into con

sideration the rounded to three significant figures)
Chemistry
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

v = 1130 cm³

Explanation:

Given data:

Volume of sample = ?

Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)

Density of Al sample = 2.70 g/cm³

Solution:

Formula:

d = m/v

d = density

m = mass

v= volume

by putting values

2.70 g/cm³ = 3057g /v

v = 3057g /2.70 g/cm³

v = 1130 cm³

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777dan777 [17]

Answer:

A. There was still 140 ml of volume available for the reaction

Explanation:

According to Avogadro's law, we have that equal volumes of all gases  contains equal number of molecules

According to the ideal gas law, we have;

The pressure exerted by a gas, P = n·R·T/V

Where;

n = The number of moles

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Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same

There will still be the same volume available for the reaction.

5 0
2 years ago
Which climate condition will cause the fastest chemical weathering of granite sandstone and shale
Naily [24]

Answer:

1. Tropical Climate

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6 0
2 years ago
Read 2 more answers
What volume of a 1.0 M HCl is required to completely neutralize 25.0 ml of a 1.0 M KOH?
kirza4 [7]
The question above can be solved by using this equation: 
CAVA =CBVB
Where:
CA =Concentration of acid = 1.0 M
VA = Volume of acid = ?
CB = Concentration of base = 1.0 M
VB = Volume of base = 25 ml
VA = CBVB / CA
VA = [1 * 25] / 1 = 25 / 1 = 25
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Therefore, the volume of acid that is required to completely neutralize the base is 25 ml.<span />
8 0
3 years ago
NEED HELP I WILL RATE BRAINLIEST!!!!
dlinn [17]
Hope this helps!! You were correct for Question 9 but incorrect for Question 10. Yes, the value comes out to be 62.4 kJ but watch out for your signs! It is definitely a positive number. Therefore, you have to consider your other choices. When I converted the 62.4 kJ to J, I found an option that matches my answer.

5 0
3 years ago
The diameter of a chlorine atom is 200. pm. How many chlorine atoms are required to line up end to end to stretch a distance of
BartSMP [9]

The number of chlorine atoms needed would simply be the ratio of distance and diameter. But first convert 200 pm to mm:

<span>200 pm = 2 E-7 mm              </span>

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7 0
2 years ago
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