Answer:
4.4 mol.
Explanation:
Hello!
In this case, since the formula for calculating the molarity is:
Whereas n stands for moles and V for the volume in liters; we can solve for n as shown below when we are given the volume and the molarity:
Thus, we plug in the given data to obtain:
Best regards!
Answer:
Kenma I'll help you! it's B.
Explanation:
I hope I helped you Kenma
Answer:
14.5 g silver
Explanation:
This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:
Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2
We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.
amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn
atomic weight of Zn: 65.37 g/mol
mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn
We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:
0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag
and the mass of silver then will be given by multiplying by the atomic weight of silver:
0.135 mol Ag x 107.9 g/mol = 14.5 g Ag
Answer:
A
Explanation:
What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3
3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply
3x = 2 * 300
3x = 600 Divide by 3
3x/3 = 600/3 Do the division
x = 200.00
