54.15 g
First you start out with the equation n=cv (n= moles, c=molarity, v= volume)
You’re going to multiply 0.10M by 3.30L to get an answer of 0.33 moles of Ca(NO3)2
From there you’re gonna convert the moles to grams to get your answer, first you have to find the molar mass of Ca(NO3)2
This can be done by finding adding the molar mass of each individual substance
The answer you should get for the molar mass is 164.1 g
From there just multiply the number of moles you calculated (0.33 mol) by the molar mass (164.1 g) and your answer is going to be 54.15 g Ca(NO3)2
1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

We can calculate the mass percent of O in MgO using the following expression.

You can learn more about mass percent here: brainly.com/question/14990953
Answer:
C. decrease the velocity.
0.0857
Rounding off to three digits
Since 7 is more than five you change it to 0 and add 1 to 5
A: 0.086
Answer:
B. A chemist
<em>Hope it's help :)</em>