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ivanzaharov [21]
3 years ago
10

How many millimeters are equivalent to 8 meters?

Chemistry
2 answers:
nataly862011 [7]3 years ago
6 0
It is the answer D. 8,000mm
Fiesta28 [93]3 years ago
3 0
D is the correct answer

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Balance the following:<br> _H2SO4 + _NaNO2 → _HNO2 + _Na2SO4
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it is balanced

Explanation:

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1. Lithium, water, edible salt, chalk, Carbon, Lime, Nitrogen, Potassium, Oxygen,
Kaylis [27]

Answer:

Explanation:

An atom is the smallest unit of an element that can take part in a chemical reaction. Atoms (and there corresponding symbols) mentioned in the question are

Lithium ⇒ Li

Carbon ⇒ C

Nitrogen ⇒ N

Potassium ⇒ K

Oxygen ⇒ O

Iron ⇒ Fe

Chlorine ⇒ Cl

A compound is substance that contains two or more atoms that are chemically combined and can be represented with a chemical formula. The compounds (and there corresponding formula) mentioned in the question are

Water ⇒ H₂O

Edible salt (sodium chloride) ⇒ NaCl

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Lime (calcium oxide) ⇒ CaO

Iodides (such as sodium iodide and potassium iodide) ⇒ NaI and KI respectively

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2 years ago
20.What do carbon and silicon have in common?
MAXImum [283]

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What provides the energy to set the water cycle in motion?
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3 years ago
A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
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