Answer: 14.01 moles
To solve this question, you need to determine the molecular mass of the Ba(OH)2. The molecular mass would be: 137.3 + 2(16+1)= 171.3g/mol.
If the total weight of Ba(OH)2 is 2400g and its molecular mass is 171.3g/mol, the number of moles would be: 2400g/ (171.3g/mol)=14.01 moles
It's measured in micrograms per deciliter. The amount is 5 micrograms.
You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.
You get: BrO3 + N2H4 -> Br2 + N2 + H2O
# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2
Br:
Multiply the reactant (left) side by 2 to balance.
O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.
H:
The product side is now equal to 12. Multiply the reactant side by 3 to balance.
N:
Now you have to balance N because the reactant side has been risen. So multiply the product side by three as well.
You end up with the complete and balanced equation:
2BrO3 + 3N2H4 -> Br2 + 3N2 + 6H2O
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;


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;


The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693.
grams
Radio waves.
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infrared waves.
light.
ultraviolet waves (rays)
x-rays.
gamma rays.