In this item, we calculate the moles of NaCl with the basis of 1L of solution.
First, calculate the mass of the solution using the density.
m = (1L) x (1000g/1000mL)(1000mL/1L) = 1000g
Then, we calculate mass of NaCl by multiplying the calculate value with the percentage.
m of NaCl = (1000 g)(0.0090) = 9 g
Then, calculate the moles of NaCl by the molar mass.
molarity = (9g)(1 mole/ 58.44 g) = 0.1539 mole / L solution or 0.1529M. Then, the answer to this item is letter C.
Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Answer:
Explanation:Remember that c is always about 300,000 km/s. Use the formula you identified to explain how the two factors in the formula are related on the electromagnetic spectrum.
The specific heat of unknown material : 3.675 J/g °C
<h3>Further explanation</h3>
Given
mass of block = 22.5 g
Q = heat = 2067 J
Δt=25 °C
Required
The specific heat
Solution
Heat can be formulated :
Q = m.c.Δt
Input the value :
2067 J = 22.5 g x c x 25 °C
c = 3.675 J/g °C