Answer:
<em>Mg </em>(<em>s</em>) + 2<em>HCI2 </em>(<em>aq</em>) → <em>MgCI2 </em>(<em>aq</em>) + <em>H2 </em>(<em>g</em>)
I think this is the correct answer I not a 100% sure if it is correct.
Explanation:
Guessing
Answer:
The answer is
<h2>2 cm/year</h2>
Explanation:
To find the rate in cm/year we must first convert 200 m into cm
1 m = 100 cm
if 1 m = 100 cm
Then 200 m = 200 × 100 = 20 ,000 cm
So the rate is
<h2>
</h2>
<u>Reduce the fraction with 10,000</u>
We have the final answer as
<h3>2 cm/year</h3>
Hope this helps you
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
Moles of Hydrogen present: 100 / 2 = 50 moles
Moles of Nitrogen present: 200 / 28 = 7.14 moles
Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles
Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.
Molar ratio of Nitrogen : Ammonia = 1 : 2
Moles of ammonia = 7.14 x 2 = 14.28 moles
