Answer:
- According to the law <br> Mass of reactants = mass of product, here <br> `underset(10 g)(CaCO_(3))rarr underset(4.4 g)(CO_(2))+underset(x)(CaO)` <br> Hence, x = 10 g - 4.4 g = 5.6 g <br> Which is mass of CaO.d
- In the first compound <br> Hydrogen = 5.93 % <br> Oxygen = `(100-5.93)% = 94.07 %` <br> In the second compound <br> Hydrogen = 11.2 % <br> Oxygen `= (100-11.2)%=88.8%` <br> In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts ...
- (The ratio of Cu combining with fixed weight of oxygen in black and red oxide is 1 : 2 respectively. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.) {Check something more in the above attachment!}
- Refer to the above attachment
Explanation:
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Answer:
Major product ethoxide ion
Explanation:
- Sodium hydride acts as a strong base towards ethanol.
- Hydride ion abstracts one proton from -OH group in ethanol to produce sodium ethoxide and hydrogen gas.
- It is an example of acid-base reaction where sodium hydride acts as a base and ethanol acts as an acid
- Structure of major organic product i.e. ethoxide ion has been shown below.
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
D= 0.10125kg/m^3
EXPLAINATION
Density =mass/volume
D= 226.8/20 x 8 x 14
D=226.8/2240
D= 0.10125kg/m^3
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