Answer:
- 40.66
- 9.91
Explanation:
For the first question:
Our theoretical compound is MR₂
1 mol of MR₂ contains 1 mol of M and 2 moles of R
Let's find out the molar mass:
9.45 g/mol + 18.12 g/mol . 2 = 45.69 g/mol
We can solve this, by an easy rule of three:
1 mol of MR₂ weighs 45.69 grams
Then, 0.89 moles may weigh 40.66 g
For the second question:
Our theoretical compound is D₂G
Let's determine the molar mass:
11.45 g/mol . 2 + 44.57 g/mol = 67.47 g/mol
1 mol of anything contains 6.02×10²³ molecules. By this definition we can say that 6.02×10²³ molecules weigh 67.47 grams. Let's solve by the rule of three:
6.02×10²³ molecules weigh 67.47 g
8.84×10²² molecules may weigh (8.84×10²² . 67.47 ) / 6.02×10²³ = 9.91 g
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.
16 g/ml
<h3>Further explanation</h3>
Given
mass = 32.40 g
volume = 2.0mL
Required
Density
Solution
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
Density formula:
ρ = density
m = mass
v = volume
Input the value :
ρ = 32.40 g / 2.0 ml
ρ = 16.2 g/ml
Rules for division: The least number of significant figures in the problem determines the number of significant figures in the answer.
32.40 = 4 sig.fig
2.0 = 2 sig fig
The answer must be 2 sig fig
So the answer = 16 g/ml
Answer:
5118.50 J
Explanation:
pΔv=nRΔT ;
therefore, ΔT=PV/nR
ΔT = (6.4×10^5)(3.2×10^(-3)/1×8.314
ΔT= 2.4633×10^2 = 246.33 K
specific heat at constant pressure is given as:
c_p = 3/2R
c_p = 12.5 J/mol K
Now, substitute in equation (1)
we know that
Q=ΔU+W ;
and
W=pΔV= 6.4×10^5×3.2×10^(-3) = 2048 J
now
ΔU=CvΔT = 12.465×246.33 =3070.50 J ;
therefore
Q=3070.50+2048= 5118.50 J
