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3 years ago

Determine which elements or compounds are missing from the following reaction equation: hydrogen + ________ → water

Chemistry

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

Oxygen

Explanation:

Water's formula is H20. Which is two hydrogen and one oxygen.

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Convert 0.02 g/mL to the unit g/L.

Vilka [71]

0,02 g/mL = 20 g/L

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3 years ago

What is the molar mass of HCl??

irakobra [83]

HCl = 1 + 35.5 => 36.5 g/mol

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7 0

3 years ago

Well, we might as well try a similar question. Magnesium metal reacts with hydrochloric acid to produce hydrogen gas and an aque

Andrei [34K]

Answer: Magnesium Mg

Explanation:

Oxidization is the process by which a substance either gains oxygen or losses electrons.

The chemical reaction of the above is denoted by,

Mg(s) + 2HCl(aq) -----> MgCl2(aq) + H2(g)

Mg went from a 0 to a +2 state which would mean that it lost electrons.

It was therefore oxidized.

Please do react or comment if you need clarification or if the answer helped you. This can help other users as well. Thank you.

5 0

3 years ago

Atomic number and mass of O

lesya692 [45]

Atomic number is 8 and atomic mass is taken as 16 amu

3 0

3 years ago

Read 2 more answers

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago