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nikklg [1K]
3 years ago
9

Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? Which of these bes

t describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen show more Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen
Chemistry
1 answer:
Pani-rosa [81]3 years ago
3 0
Nitrogen monoxide has 1 oxygen atom and
Nitrogen dioxide has 2 oxygen atoms
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What is formed called when air masses with different characteristics meet
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So, when two different air masses meet, a boundary is formed. The boundary between two air masses is called a front. Weather at a front is usually cloudy and stormy. There are four different fronts- Cold, Warm, Stationary, and Occluded.
7 0
3 years ago
A compound is 21.6% Mg, 21.4% C, and 57.0% O. What is the empirical formula of the compound? ​
nignag [31]

Answer:

Compound x = MgC_{2}O_{4}

Explanation:

Let the compound be x

Assuming we have a 100g of compound x

<u>Given the following data;</u>

Magnesium, Mg = 21.6% = 21.6g

Carbon, C = 21.4% = 21.4g

Oxygen, O = 57.0% = 57.0g

Now, we would find the amount of moles for each element.

Atomic mass of Mg = 24.30g

Atomic mass of C = 12.01g

Atomic mass of O = 16.00g

<u>Amount of moles for Mg;</u>

21.6*(1/24.30) = 0.89mol

<u>Amount of moles for C;</u>

21.4*(1/12.01) = 1.78mol

<u>Amount of moles for O;</u>

57.0*(1/16.00) = 3.56mol

We then divide by the smallest to find the ratio;

0.89/0.89 = 1 Mol of Mg

1.78/0.89 = 2 Mol of C

3.56/0.89 = 4 Mol of O

Therefore, the ratio of Mg, C and O is 1:2:4.

Compound x = MgC_{2}O_{4}

Hence, the empirical formula of the compound is MgC_{2}O_{4}

7 0
3 years ago
Ggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg
Tpy6a [65]

Answer:

gggg  good one

Explanation:

r

6 0
3 years ago
During the phosphatase experiment you will use a 1% w/v of phenolphthalein di-phosphate (PPP). How much PPP do you need to make
MrMuchimi

Answer:

To make 500 mL of a solution of PPP 1% w/v you need 5g

Explanation:

A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w):

% w/v = g of solute/ mL of solution × 100

A 1% w/v means that you have 1 g of PPP per 100 mL of solution

Thus, if you need to make 500 mL of solution you will need 5 g of PPP.

I hope it helps!

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3 years ago
OH- of a solution that has a pH of 8.57
Shkiper50 [21]

Answer:

I think its -8.75

Explanation:

Hope this helps!11

8 0
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