The best formula that represents compounds only is <span>CO2, H2O, NH3.</span>
Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
The chemical equation is
Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)
Answer:
12
Explanation:
In the right hand side of the equation, there are three compound which contains O2, which are;
Cu(NO3)2 , number of oxygen atoms =3*2 =6
2NO2, number of oxygen atoms = 2*2=4
2H2O, number of oxygen atoms =2*1=2
Total number of oxygen atoms on the right side of equation = 6+4+2 =12
the answer is d. when more solute is added
