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Natasha_Volkova [10]
3 years ago
12

All of the stars in the sky, except for one, are so far away that they look like small points of light.

Chemistry
1 answer:
kiruha [24]3 years ago
8 0

Answer:

The sun.

Explanation:

It is closer to Earth, therefore bigger to us, and not just a small point of light.

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What product, including stereochemistry, is formed when CH3OCH2CH2C≡CCH2CH(CH3)2 is treated with the following reagent: H2 (exce
Bas_tet [7]

Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.  

4 0
4 years ago
Chose all of the correct statements pure vs mixture
GarryVolchara [31]

Answer:

I think that a  pure substance is a form of matter that has a constant composition and properties that are constant throughout the sample. Mixtures are physical combinations of two or more elements and/or compounds.

Explanation:

6 0
3 years ago
The theoretical yield of ammonia in an industrial synthesis was 550 kg, but only 480 kg was obtained. What was the percentage yi
bezimeni [28]

Theoretical yield 550 kg - 100%

Actual yield 480 kg - x%


x= 480*100/550 ≈ 87.3 %


Percentage yield ≈ 87.3 %.



8 0
3 years ago
What is Suspension!? ​
aksik [14]

Answer:

A suspension is a heterogeneous mixture of a finely distributed solid in a liquid. The solid is not dissolved in the liquid, as is the case with a mixture of salt and water. Suspension.

Explanation:

Common examples of suspension include the mixture of chalk and water, muddy water, the mixture of flour and water, a mixture of dust particles and air, fog, milk of magnesia, etc. ... Particles of the suspension are large enough to scatter the rays of light and the path of ray is visible through it.

8 0
3 years ago
Read 2 more answers
Q.11. Calculate the mass of NaBr needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaC
klemol [59]

Answer:

The mass of NaBr needed is 0.22969 g.

Explanation:

1 mole of NaBr contains 22.4 dm^3 of NaBr

Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr

MW of NaBr = 23 + 80 = 103 g/mol

Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g

6 0
3 years ago
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