These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
N + O ---------> NO
Mole ratio of N to NO is 1:1
Mole of 25.0 g of NO = 25/30.01 = 0.833 moles
Equivalent mole of N = 0.833 moles
Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
55.18 L
Explanation:
First we convert 113.4 g of NO₂ into moles, using its molar mass:
- 113.4 g ÷ 46 g/mol = 2.465 mol
Then we<u> use the PV=nRT formula</u>, where:
- P = 1atm & T = 273K (This means STP)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
Input the data:
- 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K
And <u>solve for V</u>:
Answer:
The correct answer is B. Dull appearance
Explanation:
First step: convert 22.4% into fraction
that is 22.4/100 which is equivalent to 0.224
The mass of Cucl2 contained in 75.85 of 22.4% by mass of solution of CUcl2 in water is therefore,
0.224 x75.85=16.99g
alternatively
75.85 --->100%
? 22.4%
by cross multiplication =(22.4 x 75.85)/100 =16.99g
