The low temperature outside lowers the volume of the gas according to Charles' law because this law describes how a gas will behave at constant pressure. It shows that the volume of a given mass of a gas is directly proportional to the absolute temperature provided the pressure remains constant. An increase in temperature leads to an increase in volume while a decrease reduces the volume. This is due to the reduction in the distances traveled by the vibrating particles of the gas because of the lost kinetic energy.
The mixture of rock particle sand humus is called the soil.
If soil contains greater proportion of big particles it is called sandy soil. If the proportion of fine particles is relatively higher, then it is called clayey soil. If the amount of large and fine particles is about the same, then the soil is called loamy.
Answer:
Explanation:
<u>1) Data:</u>
Base: NaOH
Vb = 15.00 ml = 15.00 / 1,000 liter
Mb = ?
Acid: H₂SO₄
Va = 17.88 ml = 17.88 / 1,000 liter
Ma = 0.1053
<u>2) Chemical reaction:</u>
The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:
- Acid + Base → Salt + Water
- H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)
<u>3) Balanced chemical equation:</u>
- H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)
Placing coefficient 2 in front of NaOH and H₂O balances the equation
<u>4) Stoichiometric mole ratio:</u>
The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:
- 1 mole H₂SO₄ : 2 mole NaOH
<u>5) Calculations:</u>
a) Molarity formula: M = n / V (in liter)
⇒ n = M × V
b) Nunber of moles of acid:
- nₐ = Ma × Va = 0.1053 (17.88 / 1,000)
c) Number of moles of base, nb:
- nb = Mb × Vb = Mb × (15.00 / 1,000)
d) At equivalence point number of moles of acid = number of moles of base
- 0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)
- Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
![0.120=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=0.120%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[NH_3]](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)
Therefore, the initial rate of the reaction will be, 0.03 M/s
Answer:
The answer to your question is:
Explanation:
Data
carbon 7.3% = 7.3g
hydrogen 4.5% = 4.5g
oxygen 36.4% = 36.4 g
nitrogen 31.8% = 31.8 g
Now
For carbon
12 g --------------------1 mol
7.3 g ------------- x
x = 7.3/12 = 0.608 mol
For hydrogen
1 g -------------------- 1 mol
4.5 g ------------------ x
x = 4.5 mol
For oxygen
16 g ------------------- 1 mol
36.4 g ---------------- x
x = 2.28 mol
For nitrogen
14 g ---------------- 1 mol
31.8 g --------------- x
x = 2.27 mol
Now divide by the lowest result, the is 0.608 from carbon
carbon 0.608/0.608 = 1
hydrogen 4.5/ 0.608 = 7.4
oxygen 2.28/0.608 = 3.75
nitrogen 2.27/0.608 = 3.73
Empirical formula = CH₇O₄N₄
