Answer:
33 g.
Explanation:
Hello there!
In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

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Answer:
Both have the same amount of particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.
This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.
Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.
Answer:
The solution will turn red.
Explanation:
HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺
(red) (yellow)
Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,
At about pH 3.4, the two the forms are present in equal amounts, and the indicator colour is orange.
If you add more acid, you are disturbing the equilibrium.
According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.
The system will try to get rid of the added acid, so the position of equilibrium will move to the left.
More of the unionized molecules will form, so the solution will turn red.
<u>Answer:</u> The rate law of the reaction is ![\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5BC_2O_4%5E%7B2-%7D%5D%5E2)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:

Rate law expression for the reaction:
![\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5Ea%5BC_2O_4%5E%7B2-%7D%5D%5Eb)
where,
a = order with respect to 
b = order with respect to 
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 2 from 1, we get:

Dividing 2 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5E1%5BC_2O_4%5E%7B2-%7D%5D%5E2)
Answer: It is called a Crookes Tube, and he used it to discover cathode rays, which were later determined to be electrons.