Answer:
See explanation
Explanation:
A balanced chemical reaction equation has the same number of atoms of each element on both sides of the reaction equation.
Hence, for the reaction between KOH and H2SO4, the balanced chemical reaction equation is;
H2SO4(aq) + 2KOH(aq) ---------> K2SO4(aq) + 2H2O(l)
Complete ionic equation;
2H^+(aq) + SO4^2-(aq) + 2K^+(aq) +2OH^-(aq) -------> SO4^2-(aq) + 2K^+(aq) + 2H2O(l)
Net ionic equation;
2H^+(aq) + 2OH^-(aq) -------> 2H2O(l)
Answer:
The number of stamps and cards Maggie has left if she gives 45 stamps to a friend is 183
Explanation:
If Maggie gives 45 stamps to a friend, you must calculate the number of stamps and cards she has left.
You know Maggie has 4 folders with 30 stamps in each folder. So the number of stamps she owns is calculated as:
4 folders*30 stamps in each folder= 120 stamps
If Maggie gives 45 stamps to a friend, then the number of stamps she has left will be calculated as the difference (the subtraction) between the stamps she owned and the ones she gives away:
120 stamps - 45 stamps= 75 stamps
On the other hand, she has 3 binders with 36 baseball cards in each binder. So the number of cards she owns is calculated as:
3 binders * 36 baseball cards in each binders= 108 baseball cards
The number of stamps and cards you have left is calculated as:
75 stamps + 108 baseball cards= 183
<em><u>The number of stamps and cards Maggie has left if she gives 45 stamps to a friend is 183</u></em>
Answer:
C. 70%
Explanation:
Atomic Mass of the silicon = 28 g.
Atomic mass of the Carbon = 12 g.
Total mass of the Silicon Carbide = 28 + 12
= 40 g.
Now, Using the formula.
% Composition = Mass of the silicon/Total mass of the compound × 100 %
= 28/40 × 100 %
= 70 %
Hence, % composition of the silicon in SiC is 70%
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
