Answer:
116 g
Explanation:
From the question given above, the following data were obtained:
Number of mole of calcium = 2.9 moles
Mass of calcium =.?
The mole and mass of a substance are related according to the following formula:
Mole = mass / molar mass
With the above formula, we can obtain the mass of calcium. This can be obtained as follow:
Number of mole of calcium = 2.9 moles
Molar mass of calcium = 40 g/mol
Mass of calcium =.?
Mole = mass / molar mass
2.9 = mass of calcium / 40
Cross multiply
Mass of calcium = 2.9 × 40
Mass of calcium = 116 g
Therefore, the mass of 2.9 moles of calcium is 116 g.
Answer:
Noble gas Electronic configuration of arsenic:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Explanation:
Arsenic is metalloid.
Its atomic number is 33.
Its atomic mass is 75 amu.
Its symbol is As.
It is usually present in combine with sulfur and metals.
it is used in bronzing.
It is also used for hardening.
Electronic configuration:
As₃₃ = Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³
Noble gas Electronic configuration:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Noble gas electronic configuration is shortest electronic configuration by using the noble gas elements full octet electronic configuration.
Answer:
Mass = 157.5 g
Explanation:
Given data:
Mass of CO needed = ?
Mass of Fe formed = 209.7 g
Solution:
Chemical equation:
3CO + F₂O₃ → 2Fe + 3CO₂
Number of moles of Fe:
Number of moles = mass/ molar mass
Number of moles = 209.7 g/ 55.85 g/mol
Number of moles = 3.75 mol
Now we will compare the moles of iron and carbon monoxide.
Fe : CO
2 : 3
3.75 ; 3/2×3.75 = 5.625 mol
Mass of CO:
Mass = number of moles × molar mass
Mass = 5.625 mol × 28 g/mol
Mass = 157.5 g
Answer:
B should be the answer, and ur low-key valid lol
Explanation:
We can use combined gas laws to solve for the volume of the gas
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L
