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qwelly [4]
3 years ago
8

An automobile can travel 40.0 miles on one gallon of gasoline how many kilometers per liter is this

Chemistry
2 answers:
Dmitrij [34]3 years ago
8 0
Sent a pic of your problem and worked it out.

Leya [2.2K]3 years ago
6 0

Answer: 17.04 km/L

Explanation:

Given: An automobile can travel 40.0 miles on one gallon of gasoline

That is 40 miles/ gallon

Now using the conversion factor :

1 mile = 1.61 km

Thus 40 miles =\frac{1.61}{1}\times 40=64.4km

Alsousing the conversion factor

1 gallon = 3.78 L

Thus 40miles/gallon will be \frac{64.4km}{3.78L}=17.04km/L

Thus the answer is 17.04 km/L.

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Small, highly charged metal cations may interact with H2O to give solutions that are not neutral. Which net ionic chemical equat
Vaselesa [24]

Answer:

Option B is correct.

The net ionic chemical reaction when Cu(NO₃)₂ reacts with water is

Cu²⁺(aq) + H₂O(l) ⇌ Cu(OH)¹⁺(aq) + H¹⁺(aq)

Explanation:

When ions dissolve in polar solvents (in water especially), then often attract ions of opposite signs.

Cu(NO₃)₂ dissolves into Cu²⁺ & NO₃⁻ and water contains H¹⁺ & OH⁻

The Cu²⁺ attracts the OH⁻ from water, thereby giving the Cu(OH)¹⁺ ion (with a net charge of +2-1 = +1) and the NO₃⁻ takes up the H⁺ ion.

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3 years ago
2. What are some strengths and limitations of the constellations
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3 0
3 years ago
Read 2 more answers
Which of the following is an example of an environmental problem caused by chemicals?
Dmitrij [34]

Answer: c

Explanation:

3 0
3 years ago
I need help with this packet
Anna11 [10]
Q1. An inorganic compound is a compound where the main constituent or substance is not that of Carbon but predominantly other elements, such as I, N etc. An organic compound is one where the main substituent or main element, the element found in much greater amounts would be Carbon.

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5 0
3 years ago
Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p
Oduvanchick [21]

Answer:

Procedure (2)  

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

\dfrac{5}{4000} = \dfrac{1}{800}

Procedure (2)

For the first dialysis, the factor is

\dfrac{5}{1000} = \dfrac{1}{200}

After a second dialysis, the original concentration of NaCl will be reduced by a factor of  

\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}

Procedure (2) is more efficient by a factor of  

\dfrac{40000}{800} = \mathbf{50}

4 0
3 years ago
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