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Ann [662]
3 years ago
6

Which diagram shows how CH3Br and OH must collide in order to react and form CH2OH? (Marking brainliest!)

Chemistry
2 answers:
makkiz [27]3 years ago
7 0

Answer:

Its B

Explanation:

I just took the quiz.

vodka [1.7K]3 years ago
4 0

Answer:

did u get the answer

sjsj

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Which of these are broken down during chemical,but not physical,changes
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Which of these what?

This isn't clear, can you put the answer choice or something?

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1.34 milligrams is the same as _______kg and ______g
Maurinko [17]
There are 1000 mg in 1 g
and there are 1000 g in 1 kg

Start by converting 1.34 mg to grams by dividing 1.34 mg by 1000 g = 0.00134 g

Then convert 0.00134 g to kg by dividing 0.00134 g by 1000 kg = 1.34×10^-6 kg OR 0.00000134 kg
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Which of the following uses a nonrenewable energy resource to generate energy? A. nuclear power plant B. wind farm C. fireplace
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What is the mass of KNO3 should be used to prepare 2.00 L of a 0.700 M solution?
Dmitry [639]

Answer:

141.4g

Explanation:

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3 0
3 years ago
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
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