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Answer:
The number of mol is: 0, 042 mol in 4 grams of MgCl2
Explanation:
We calculate the weight of 1 mol of MgCl2:
Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=
24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2
95, 3 grams------1 mol MgCl2
4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2
Answer: I am confident the answer is B
Explanation:
forgive me if im wrong
Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
Answer:
0.052mL
Explanation:
1mole of a gas occupy 22.4L.
Therefore, 1 mole of CO2 will also occupy 22.4L.
If 1mole of CO2 occupies 22.4L,
Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L
coverting this volume to mL, we simply divide by 1000 as shown below:
51.52/1000 = 0.05152mL = 0.052mL