<span>A.to calculate the amount of product that would form
</span>
Answer:
Part 1) 85.3 grams NaCl
Part 2) 8.79 x 10²³ formula units NaCl
Explanation:
<u>(Part 1)</u>
To find the mass of NaCl, you need to multiply the given value (1.46 moles) by the molar mass of NaCl. This measurement is the atomic masses of the elements times each of their quantities combined. In this case, there is only one mole of each element in the molecule. Moles should be located in the denominator of the conversion to allow for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
1.46 moles NaCl 58.44 g
--------------------------- x ---------------- = 85.3 grams NaCl
1 mole
<u>(Part 2)</u>
I do not know which other question the second part is referring to, so I will just use the moles given in the first part. To find the formula units, you need to multiply the given value (1.46 moles NaCl) by Avogadro's Number. This conversion represents the number of formula units found in 1 mole of the sample. The moles should be in the denominator of the conversion to allow for the cancellation of units.
Avogadro's Number:
1 mole = 6.022 x 10²³ formula units
1.46 moles NaCl 6.022 x 10²³ units
------------------------ x ----------------------------- = 8.79 x 10²³ formula units NaCl
1 mole
Answer:k
Explanation:
Atomic theory would be the theory or explanation of the function of atoms and their relationship with each other.
Answer:
56972.17K
Explanation:
P = 4.06kPa = 4.06×10³Pa
V = 14L
n = 0.12 moles
R = 8.314J/Mol.K
T = ?
We need ideal gas equation to solve this question
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
T = PV / nR
T = (4.06×10³ × 14) / (0.12 × 8.314)
T = 56840 / 0.99768
T = 56972.17K
Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa
The answer is B. Cellular Respiration, for sure.
