To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M
So, the answer to 27.) would be <em>2x.</em> Both 6x and 2x can be divided by 2x, but they can't go any higher without the end-answer becoming a fraction. As such, 2x is the greatest common factor.
For 28.), x and x^2 can't be like terms, since like terms have the same variable and exponent :)
Hope I could help!
Answer:
See explanation
Explanation:
The molecular equation shows all the compounds involved in the reaction.
The molecular equation is as follows;
2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)
The complete ionic equation shows all the ions involved in the reaction
The complete ionic equation;
2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)
The net Ionic equation shows the ions that actually participated in the reaction
The net ionic equation is;
2F^-(aq) + Pb^2+(aq)--------> PbF(s)
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).