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3 years ago

Hey conner i miss you soooooo much and does anyone have tik tok

Chemistry

2 answers:

irakobra [83]3 years ago

8 0

Answer:

i have tik tok

Explanation:

kherson [118]3 years ago

6 0

Answer:

I have tik tok

Explanation:

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5. __NH3 + __O2 >>>>>__ NO +__ H20

inn [45]

2 NH3+ 2 O2 —> 2 NO+ 3 H2O

5 0

3 years ago

Fied the emicicol formula of a compoun containing<br> 2.44% hydrogen, 39.02% sulphur, 58.54% oxygen

Viefleur [7K]

The answer is in the attachment below:

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2 years ago

When magnesium loses it's valence electrons, it has the atomic structure of what element?

Arte-miy333 [17]

As far as I can tell the best answer for this would be (A) Neon. However, I would argue that this is at the very least a misleading question. Atoms are less identified by their electrons than their protons (which is represented always by its atomic number). Although atoms can gain or lose electrons, the protons would never change (and remain the same element). Personally, I would have written the question as, "When Magnesium loses its valence electrons, its new number of electrons would most closely resemble _____"

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3 years ago

Assume a density of water of 1.00 g/mL, and calculate the mass of water in the solution

tiny-mole [99]

Answer:

1g or 10^-3kg

Explanation:

as you know , the density =mass \volume

so you have the mass from the number it self

<h2> $\frac{1g}{1 ml}$ so from this equation, you will get 1 g and you can to SI to be $10^{-3}$ kg</h2>

5 0

3 years ago

A sample of an unknown gas takes 371 s to diffuse through a porous plug at a given temperature.

blsea [12.9K]

Answer:

125.84 g/mol is the molar mass of the unknown gas.

Explanation:

Let the volume of the gases effusing out be V.

Effusion rate of the unknown gas = $R=\frac{V}{371 s}$

Effusion rate of the nitrogen gas = $r=\frac{V}{175 s}$

Molar mass of unknown gas = m

Mass of nitrogen gas = 28 g/mol

Graham's law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{R}{r}=\sqrt{\frac{28 g/mol}{M}}$

$\frac{\frac{V}{371 s}}{\frac{V}{175s}}=\sqrt{\frac{28 g/mol}{M}}$

$\frac{175 s}{371 s}=\sqrt{\frac{28 g/mol}{M}}$

$M=\frac{28 g/mol\times 371\times 371}{175\times 175}=125.84 g/mol$

125.84 g/mol is the molar mass of the unknown gas.

5 0

3 years ago