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docker41 [41]
2 years ago
14

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade​

Chemistry
1 answer:
egoroff_w [7]2 years ago
5 0

Answer:

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

Explanation:

Assuming Pressure is constant.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

\frac{300}{273} = \frac{V_2}{288}

V_2 = \frac{288*300}{273} \\V_2 = 316.5

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

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la longitud de un cable de aluminio es de 30 m a 20 C. Sabiendo que el cable es calentado hasta 60 C y que el coeficiente de dil
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Answer:

30cm

Explanation:

6 0
3 years ago
Which of the following statements best compares the pH value of H3PO4 and HCl?
wel

Answer:

Option B. Both have a pH less than 7, but H3PO4 has a lower pH than HCl

Explanation:

Those are acid, so the pH would be < 7.

H₃PO₄ is a weak acid with 3 dissociations

HCl is a strong acid.

pH depends on [H]⁺

H₃PO₄  →  3H⁺  +  PO₄⁻³

HCl →  H⁺  +  Cl⁻

If both acid, have the same concentration, [H⁺]H₃PO₄ > [H⁺]HCl, that's why the pH from the phosphoric will be lower.

5 0
3 years ago
If the pressure of a gas is 1.01 atm and the temperature is 25°C, then if the pressure is increased to 1.10, what is the new tem
vodka [1.7K]

Answer:

T2 = 51.6°C

Explanation:

Given:

P1 = 1.01 atm

T1 = 25°C + 273 = 298K

P2 = 1.10 atm

T2 = ?

P1/T1 = P2/T2

Solving for T2,

T2 = (P2/P1)T1

= (1.10 atm/1.01 atm)(298K)

= 324.6 K

= 51.6°C

where Tc = Tk - 273

3 0
3 years ago
Please help me! During the replication phase, DNA:
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3 years ago
Read 2 more answers
How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide
Marat540 [252]

Answer:

\boxed{\text{2.00 mol}}

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r:      58.32

          Mg(OH)₂ + … ⟶ … + 2HOH

m/g:       58.3

(a) Moles of Mg(OH)₂

\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}

The reaction will form \boxed{\textbf{2.00 mol}} of water.

6 0
3 years ago
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