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Arturiano [62]
3 years ago
5

Which of the following is not a step in the conversion of 1.3 dam3 to L?

Chemistry
1 answer:
Tanya [424]3 years ago
3 0
First put the choices,and the answer is <span>(10^-3L/1mL)^3 is the correct answer.</span>
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PLS HELP
Juli2301 [7.4K]

Answer:

False

Explanation:

Water molecules are spread up but slightly, enough that none of the molecules actually move freely around each other. Only gas and plasma molecules are very spread apart.

7 0
2 years ago
Write the formulas of the compounds formed by specified cations with the following anions: OH-, ClO3-, SO3^2-, HPO4^2-. . . . Ex
Katena32 [7]
In the follow

<span>1)N<span>H4</span>OH,N<span>H4</span>Cl<span>O3</span>,(N<span>H4</span><span>)2</span>S<span>O3</span>,(N<span>H4</span><span>)2</span>HP<span>O4</span></span> <span>2)Al(OH<span>)3</span>,Al(Cl<span>O3</span><span>)3</span>,A<span>l2</span>(S<span>O3</span><span>)3</span>,A<span>l2</span>(HP<span>O4</span><span>)3</span></span> <span><span>3)Pb(OH<span>)4</span>,Pb(Cl<span>O3</span><span>)4</span>,Pb(S<span>O3</span><span>)2</span>,Pb(HP<span>O4</span><span>)2</span></span></span>
7 0
3 years ago
Read 2 more answers
Hugh wrote the properties of physical and chemical weathering in the table shown.
Talja [164]

Answer:

I am sure that the C one is correct

6 0
3 years ago
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What is the purpose of the uninoculated control tubes used in the oxidation fermentation test?
Black_prince [1.1K]
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions. 
5 0
3 years ago
At a certain temperature the vapor pressure of pure heptane C7H16 is measured to be 454.mmHg. Suppose a solution is prepared by
OlgaM077 [116]

Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

No of mole of heptane = 1.0179

Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

Partial pressure of heptane = 1.0179*454mmHg

Partial pressure of heptane = 462.1096 = 462mmHg

the partial pressure of heptane vapor above this solution = 462mmHg

5 0
3 years ago
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