the answer is d hope this helps
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, in terms of pressures, the rate becomes:

Thus, the rate of change for the partial pressure of ammonia turns out:
![r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h](https://tex.z-dn.net/?f=r_%7BNH_3%7D%3D2%2A%28-r_%7BN_2H_4%7D%29%5C%5Cr_%7BNH_3%7D%3D2%2A%5B-%28-70torr%2Fh%29%5D%5C%5Cr_%7BNH_3%7D%3D140torr%2Fh)
The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.
Best regards.
Hey there!
In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage.
→ Na₂CO₃ ×<span> 10 H</span>₂<span>O
</span>→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15
Now, just add all of those numbers up for the total molar mass.
→ 45.9796 + 12.0107 + 47.997 + 180.15 = <span>286.1373
</span>
The last step is to divide the molar mass of the 10 water molecules by the total mass.
→ 180.15 ÷ 286.1373 = <span>0.62959 </span>≈ 0.63
Your answer will be about 63%.
Hope this helped you out! :-)

We know, 1 m³ of space can hold 1000 l of the substance.
⇛ 1 m³ = 1000 l----(1)
And, 1 l is 1000 times more than 1 ml
⇛ 1 l = 1000 ml------(2)
So, From (1) and (2),
⇛ 1 m³ = 1000 × 1000 ml
⇛ 1m³ = 1000000 ml
We had to find,
⇛ 1.40 m³ = 1.40 × 1000000 ml
⇛ 1.40 m³ = 140/100 × 1000000 ml
⇛ 1.40 m³ = 1400000 ml
⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml
☃️ <u>So</u><u>,</u><u> </u><u>1.40</u><u> </u><u>m</u><u>³</u><u> </u><u>=</u><u> </u><u>1</u><u>4</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>⁵</u><u> </u><u>m</u><u>l</u><u> </u><u>/</u><u> </u><u>1.4</u><u> </u><u>×</u><u> </u><u>10</u><u>⁶</u><u> </u><u>ml</u><u>.</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>