Answer:
The mass percent of copper as element is the same.
Explanation:
First of all we need the reaction that is presented below:
→ 
The mass percent of copper (Cu) as element is the same because of during the reaction the element only transform its nature from copper carbonate (
) to copper oxide (
), the latter is a solid and will remain in the system.
On the other hand, you will note that the global percentage mass will be small because of the reaction produce (
) that is a gas and this one will escape for the system.
Have a great day!
Answer:
See explanation
Explanation:
The magnitude of electronegativity difference between atoms in a bond determines whether that bond will be polar or not.
If the electronegativity difference between atoms in a bond is about 1.7, the bond is ionic. If the electronegativity difference is greater than 0.4 and less than 1.7, the bond will have a polar covalent character. Lastly, if the electronegativity difference between the bond is less than or equal to 0.4, the covalent bond is non polar.
The electronegativity difference between carbon and hydrogen is about 0.4 which corresponds to a nonpolar covalent bond hence the molecule is nonpolar.
The electronegativity difference between carbon and fluorine is about 1.5 indicating a highly polar bond. This gives CH3F an overall dipole moment thereby making the molecule polar.
Heat
gained or loss in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC(T2-T1)
When two objects are in contact,
it should be that the heat lost is equal to what is gained by the other. So, the heat released by the lead is equal to the heat that is absorbed by the water.
</span>Heat = mC(T2-T1) = 50.0 mL (1.00 g/mL) (4.18 J/g °C) (20 °C - 18 °C) = 418 J<span>
</span>
Answer:
It is traveling at 170 miles per hour.
Explanation:
divide 340 by 2 because right now it's at 340 miles per two hours.
You get 170 miles per hour.
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams