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4 years ago

As the pressure on a sample of gas increases, the volume of the sample _____.

Chemistry

2 answers:

Free_Kalibri [48]4 years ago

7 0

The volume of the sample increases.

ki77a [65]4 years ago

4 0

Answer:

The volume decreases

Explanation:

According to Boyles law the relationship is inverse

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Scientific evidence is most likely to be consistent if it is based on data from

ladessa [460]

Answer:

Random samples

Explanation:

It needs to be random so that there isn't bias that would skew the consistency

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4 years ago

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Please help! This is for a test and I want to have a good grade. Have a great day! :)

elena-s [515]

Answer:

genetic engineering !! :)

Explanation:

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3 years ago

Why must contact lenses be removed before you enter a chemistry laboratory?

RUDIKE [14]

The vapors of some chemicals used in the chemistry laboratory, especially isocyanates, can react and bond the contact lens to the eye.

(FYI - the family of isocyanates include Superglue..I wouldn't want to have been the person who discovered this unfortunate reaction!)

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4 years ago

Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil

emmainna [20.7K]

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

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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