2.3 mols of Al*26.98 molar mass of Al=62.054g
this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
It would be C
2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj
Answer:
Option D. 3, 1, 3, 1
Explanation:
From the question given above,
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
The equation can be balance as follow:
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:
3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:
3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃
Now, the equation is balanced.
Thus, the coefficients are 3, 1, 3, 1
