Answer: n∗R=22+273.15/4.2∗5n
P2=n∗R∗T2/V2=n∗R∗33.6+273.15/10
Explanation:
Explanation:
there is 2 nitrogen but if you mean nitrate is 6
Answer:
c : 13%
Explanation:
Data Give:
Experimental density of vanadium = 6.9 g/cm³
percent error = ?
Solution:
Formula used to calculate % error
% error = [experimental value -accepted value/accepted value] x 100
The reported accepted density value for vanadium = 6.11 g/cm³
Put value in the above equation
% error = [ 6.9 - 6.11 / 6.11 ] x 100
% error = [ 0.79 / 6.11 ] x 100
% error = [ 0.129] x 100
% error = 12.9
Round to the 2 significant figure
% error = 13 %
So, option c is correct
M C C O R M I C K T E A M ’ S O P E N Q U A N T U M M AT E R I A L S D ATA B A S E
O F F E R S U N L I M I T E D A C C E S S T O
ANALYSES OF NEARLY 300,000 COMPOUNDS
Answer:
a. 53.5 g/mol
b. 80.06 g/mol
c. 133.33 g/mol
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Molar Mass - 1 mol per <em>x</em> grams substance
Explanation:
<u>Step 1: Define</u>
a. NH₄Cl
b. NH₄NO₃
c. AlCl₃
<u>Step 2: Find masses</u>
Molar Mass of N - 14.01 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of Cl - 35.45 g/mol
<u>Step 3: Calculate compound masses</u>
Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol
Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol
Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol
