Answer:
The oxidation number of C (carbon) is +4
Explanation:
I think it may be gravity. Are there answer choices??
To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.
We want to standardize a H₂SO₄ solution with NaOH. The neutralization reaction is:
H₂SO₄ + 2 NaOH ⇒ Na₂SO₄ + H₂O
The buret, which contains NaOH, reads initially 0.63 mL, and 44.73 mL at the endpoint. The used volume of NaOH is:

44.10 mL of 0.450 M NaOH are used for the titration. The reacting moles of NaOH are:

The molar ratio of H₂SO₄ to NaOH is 1:2. The moles of H₂SO₄ that react with 0.0198 moles of NaOH are:

0.00990 moles of H₂SO₄ are in 20.00 mL of solution. The molarity of H₂SO₄ is:
![[H_2SO_4] = \frac{0.00990mol}{0.02000} = 0.495 M](https://tex.z-dn.net/?f=%5BH_2SO_4%5D%20%3D%20%5Cfrac%7B0.00990mol%7D%7B0.02000%7D%20%3D%200.495%20M)
To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.
Learn more: brainly.com/question/2728613
Answer:
Mass of 1 litre of this gas at 273 degree Celsius and 1140 mm Hg pressure is
3 grams. So, (T1) = (273 + 273) degree Absolute = 546 degree Absolute and
(P1) = (760 + 1140) mm Hg = 1900 mm Hg and (V1) = 1 Litre
First, we have to find out how much volume does it occupy at NTP; (T2) = 273 degree Absolute and (P2) = 760 mm Hg.
So, (V2) = (1900*1 / 546)*(273 / 760) = 1.25 litre.
So, the mass of 1.25 litre volume of this gas = 3 grams
Therefore, the mass of 22.4 litre volume of this gas = (3*22.4 / 1.25) grams
= 53.76 grams.
So, the gram molecular mass of the gas is 53.76.
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Answer:
I think that the irregular shape is more dense than the cube. I am not completely sure though