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konstantin123 [22]
3 years ago
10

Before you can find the mass, what do you need to know?

Chemistry
2 answers:
Oxana [17]3 years ago
5 0

Answer:

you need to know what you're finding the mass of.

Explanation:

This is just one example. There are probably more out there, but this is one thing my parents taught me. Hope this helps, and tell me if I'm wrong!

Dmitrij [34]3 years ago
3 0

Answer:The easiest way to find the mass of anything is to weigh it. You're actually measuring the force of gravity on the object, and technically, you should divide the weight by the acceleration due to gravity to get the mass.

Explanation:

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Determine the grams of Iron(III) chloride produced if 22.5 g Iron reacts with Chlorine gas. First, balance the chemical equation
TEA [102]
22.5 because it still stays the same
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3 years ago
Which of these steps is most likely to be part of an investigation about friction?
Serggg [28]

Answer:

I want to say option C: Testing which surfa e is easier to slide a wooden block across.

Explanation:

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4 0
2 years ago
How are objects charged by friction
mote1985 [20]

Answer:

When insulating materials rub against each other, they may become electrically charged . Electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. The material that gains electrons becomes negatively charged

Explanation:

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4 0
3 years ago
Calculate the number of moles of a gas that is present in a 7.55 L container at 45°C, if the gas exerts a pressure of 725mm Hg.
vovikov84 [41]

<u>Answer:</u> The number of moles of gas present is 0.276 moles

<u>Explanation:</u>

To calculate the number of moles of gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of the gas = 725 mm Hg

V = Volume of the gas = 7.55 L

n = number of moles of gas = ?

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of the gas = 45^oC=(45+273)K=318K

Putting values in above equation, we get:

725mmHg\times 7.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 318K\\\\n=0.276mol

Hence, the number of moles of gas present is 0.276 moles

4 0
3 years ago
Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a
densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0
3 years ago
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