590 mL = 590 cm³= 0,59 dm³
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃
:)
The red is an oxygen molecule. 1 oxygen and 2 hydrogen =h2o or water!
Answer is: there are 3·10²³ atoms of carbon-12.
m(C) = 6.00 g; mass of carbon.
M(C) = 12 g/mol.
n(C) = m(C) ÷ M(C).
n(C) = 6 g ÷ 12 g/mol.
n(C) = 0.5 mol.
N(C) = Na · n(C).
n(C) = 6·10²³ 1/mol · 0.5 mol.
n(C) = 3·10²³.
n - amount of substance.
M - molar mass.
Na - Avogadro number.
2SO2 + O2 ------> 2SO3
1) M(SO2)= 32.0 + 2*16.0 = 64 g/mol
2) 100.0 g SO2 * 1 mol SO2/64 g SO2 = 1.5625 mol SO2
3) 2SO2 + O2 ------> 2SO3
2 mol 1 mol
1.5625 mol x mol
x= 1.5625/2=0.78125 ≈ 0.7813 mol O2
Answer: 0.7813 mol O2.
Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
In such a way, solving for the final temperature we obtain:
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
Thereby, answer is B. 58°C
.
Regards.