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elena-14-01-66 [18.8K]
3 years ago
11

How many oxygen atoms are there in the following compound: 2HNO3 3 5 6 12

Chemistry
2 answers:
Fudgin [204]3 years ago
7 0

Answer:

6

Explanation:

There is 2 of the coumpond, in one compound, there are 3 oxygens. But when there are two, it is 6.

Ahat [919]3 years ago
3 0

Answer:

The answer is there are 3 oxygen atoms.

Explanation:

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48 mph

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muminat

Answer: for question 4, the answer would most likely be movement of air, water and rock

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What is the BEST definition for friction?
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4 0
3 years ago
Calculate the concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58.
Artyom0805 [142]

The concentration (in M) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

pOH is the measure of basic nature of a solution by evaluating the [OH⁻] concentration.

pH is measure of acidic nature of a solution by evaluating the [H⁺] concentration.  It is the negative logarithm of the hydroxide ion concentration. It gives hydronium ions on dissociation.

Given,

pOH = 3.58

Temperature = 25°C = 298K

We know that pH + pOH =14

pH + 3.58 = 14

pH = 10.42

pH is the negative logarithm of hydrogen ion concentration.

At 25°C, the relation of pH and [H⁺] concentration is as follows:

∴ pH = -log [H⁺]

⇒ 10.42 = -log [H⁺]

⇒log [H⁺] = -10.42

⇒ [H⁺] = antilog (-10.42)

⇒[H⁺] =3.8× 10⁻¹¹ M

The concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

Learn more about pH here, brainly.com/question/17144456

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8 0
2 years ago
A 1.018 g sample pure platinum metal was reacted with HCl to form 1.778 g of a compound containing only platinum and chlorine. D
Ivanshal [37]

Answer:

PtCl_4

Explanation:

Hello!

In this case, since HCl and Pt react according to the following chemical equation:

HCl+Pt\rightarrow PtCl_x+H_2

Whereas PtClx is the compound containing Pt and Cl; thus, since 1.018 g out of 1.778 g correspond to Pt and therefore 0.760 g to chlorine, so we determine the empirical formula of this compound by firstly computing the moles of each element:

n_{Pt}=1.018gPt*\frac{1molPt}{195.084gPt}=0.00522molPt\\\\\\n_{Cl}=0.760gCl*\frac{1molCl}{35.45gCl}  =0.0214molCl

Now, we divide the each moles by those of Pt as the fewest ones in order to compute their subscripts in the empirical formula:

Pt=\frac{0.00522}{0.00522}=1 \\\\Cl=\frac{0.0214}{0.00522} =4

Thus, the required formula is:

PtCl_4

Best regards!

3 0
3 years ago
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