Answer:
5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃
6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693.
grams of N₂
Explanation:
5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;
(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O
One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂
The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;
![\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%20%5C%20g%7D%7B105.99%20%5C%20g%7D%20%3D%20%5Cdfrac%7B5%20%5C%20g%7D%7B44.01%20%5C%20g%7D)
![\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g](https://tex.z-dn.net/?f=%5Ctherefore%20%7Bx%20%5C%20g%7D%20%3D%20%5Cdfrac%7B5%20%5C%20g%7D%7B44.01%20%5C%20g%7D%20%5Ctimes%20105.99%20%5C%20g%20%5Capprox%2012.04%20%5C%20g)
The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g
6. The chemical equation for the reaction is presented as follows;
N₂ + 3H₂ → 2NH₃
Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)
The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;
![\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%20%5C%20g%7D%7B28%20.01%20%5C%20g%7D%20%3D%20%5Cdfrac%7B150%20%5C%20g%7D%7B3%20%5Ctimes%202.02%20%5C%20g%7D%20%3D%20%5Cdfrac%7B150%20%5C%20g%7D%7B6.06%20%5C%20g%7D)
![\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g](https://tex.z-dn.net/?f=%5Ctherefore%20%5C%20x%20%5C%20g%3D%20%5Cdfrac%7B150%20%5C%20g%7D%7B6.06%20%5C%20g%7D%20%5Ctimes%2028.01%20%5C%20g%20%3D%20693.%5Coverline%20%7B3168%7D%20%5C%20g)
The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693.
grams
Waxing crescent I know that bc I took the test
Answer:
5g ($120)
Explanation:
The amount of chemical needed for assay is 3 ml of 250 mM solution of a chemical
Molarity (M) = number of moles / volume in L
number of mole = M × volume in L
M of the chemical = 250 mM = 250 / 1000 = 0.25 M
number of moles = 0.25 × ( 3 / 1000) in L = 0.00075 m
mass of the chemical needed = 0.00075 × molar mass = 0.00075 × 156 = 0.117 g for each assay.
mass needed for 40 assay = 40 × 0.117 = 4.68 g
It is therefore wise for him to buy the 5 g ( $ 120), though the 10 g and 25 g yields better prices per gram, they much more than what he needed for the assay.
That person was right it is A