Answer:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce 
= 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g
0.0381g/mL will be concentration of this solution
The ratio of a solute—a substance that dissolves—to a solvent—a substance that does not dissolve—determines the concentration of a solution in chemistry. C = m/V, where C is the concentration, m is the mass of the solute dissolved, and V is the overall volume of the solution, is the accepted formula.
How to reach the solution?
First find for the density of NaOh which is 2.13 g/cm3
3.88 ÷ 2.13 = 1.82mL
100mL + 1.82mL = 101.82mL
3.88 ÷ 101.82mL = 0.0381g/mL
More about NaOh:
The extremely adaptable chemical sodium hydroxide (NaOH), sometimes referred to as caustic soda or lye, is employed in a range of manufacturing processes. A byproduct of the manufacturing of chlorine is sodium hydroxide. Numerous items that are used on a daily basis, including paper, metal, drain and oven cleansers for industry, soap, and detergents, are made with sodium hydroxide.
Learn more about Naoh here:
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The terms of a equation, the momentum of an object is equal to the mass of the object times the velocity of the object. where m is the mass and v is the velocity
Answer:
92.37°C
Explanation:
Assuming the volume remains the same in both the states ( bicycle tire).
Then as per ideal gas equation
where P1, P2 are pressure at two different states and T1, T2 are temperature at these two states
Given:
P1 = 1.34 atm, T1 = 33.0 ° C = 306K
P2 = 1.60 atm, T2 = ?
1.34/306 = 1.60/T_2
⇒ T2 = 1.60×306/1.34
= T2= 365.37K= 92.37°C
7. a. I
8. c. 75 g/ml
9. b. .25g/cm3
10. b. the density decreases
