It's either B or D because there would be 2 coefficients of hydrogen and one of oxygen
wherears one hydrogen atom has a mass of approximately 1 u ,1 mol of h atoms has a mass of approximately 1 gram and wherears one sodium atom has an approximately 1 gram and one sodium atom have approximate mass of 23 u
Answer:
106.25 mL
Explanation:
For this, we can use
C1×V1=C2×V2
C1 = 0.45
V1 = 85
C2= 0.20
V2= ?
0.45 × 85 = 0.20 × V2
V2= (0.45 × 85)/0.20
V2=191.25mL
To find the amount of water added, subtract V1 from V2
191.25 - 85 =106.25mL
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Answer:
Approximately 
(approximately 
.)
Explanation:
Calculate the quantity 
of lithium phosphate in 
of this
lithium phosphate solution.
.
Empirical formula of lithium phosphate: 
.
Look up the relative atomic mass of 
, 
,and 
on a modern periodic table:
Calculate the formula mass of :
.
Calculate the mass of that 
of formula units:
.
.
.
.