Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
Answer:
38.152 g NaCl would be produced.
Explanation:
<u>M</u><u>a</u><u>r</u><u>k</u><u> </u><u>me</u><u> </u><u>as</u><u> </u><u>Brainliest</u><u> </u>
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
Answer:
Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Explanation:
Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
This is how it starts out.
Left:
Right
So the place to start with this equation is to bring the Cls up to 2
Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
But the Nas are now out of kilter.
Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
Now the right has a problem. There's only 1 Na
Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Check it out. It looks like we are done.
