Answer:

Explanation:
There are no molecules in NaCl, because it consists only of ions.
However, we can calculate the number of formula units (FU) of NaCl.
Step 1. Calculate the moles of NaCl

Step 2. Convert moles to formula units

There are
in 3.6 g of NaCl.
Answer:
The correct answer is - option D. (check image)
Explanation:
Alkynes and alkenes both decolorized bromine in carbon tetrachloride. The absorption of the IR at about 3300 cm-1 for the X here that are found in the terminal alkynes absorption range only. In presence of excess hydrogen and a nickel catalyst, x gives the 2-methyl pentane.
The most likely structure for X is: CH3-CH3-ch-CH2-C≡CH
Answer:
46.6 laps
Explanation:
Step 1: Given data
- Length of the lap (l): 1056 ft
- Total distance to be run (d): 15.0 km
Step 2: Convert "l" to km
We will use the conversion factor 1 km = 3280.84 ft.
1056 ft × (1 km / 3280.84 ft) = 0.3219 km
Step 3: Calculate the number of laps that you should run
We will use the following expression.
laps = d/l
laps = 15.0 km/(0.3219 km/lap)
laps = 46.6 laps
<span>carbon = 42.1%
hydrogen = 6.5%
oxygen = 51.4%
First lookup the molar mass of carbon, hydrogen, and oxygen.
mass of carbon = 12.0107
mass of hydrogen = 1.00794
mass of oxygen = 15.999
Now calculate the molar mass of each element in sucrose multiplying the atomic weight of each element by the number of times the element is used.
carbon = 12 * 12.0107 = 144.1284
hydrogen = 22 * 1.00794 = 22.17468
oxygen = 11 * 15.999 = 175.989
Calculate the molar mass of sucrose by adding the mass of each element used.
144.1284 + 22.17468 + 175.989 = 342.2921
Finally, calculate the percentage by mass of each element by dividing the mass used for each element by the total mass of sucrose.
carbon = 144.1284 / 342.2921 = 0.421068 = 42.1%
hydrogen = 22.17468 / 342.2921 = 0.064783 = 6.5%
oxygen = 175.989 / 342.2921 = 0.514149 = 51.4%</span>
Answer:
x = 4.17
y = 1.86
Explanation:
0.62 = log(x)
x = 10^0.62 = 4.17 ( to the nearest hundredth)
0.62 = ln(y)
y = e^0.62 = 1.86 (to the nearest hundredth)