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2 years ago

What is the pressure of the ideal gas when the volume is decreased to 4.0 L and the temperature is increased to 265 K?

1 answer:

8 0

It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT. First, we solve the amount of the gas in moles using the said equation and the first conditions.

(2.0 atm) (5.0 x 10^3 cm^3) = n (82.0575 atm.cm^3/mol.K)(215 K)
n=0.5668 mol

Using the second conditions given, we obtain the new pressure.
P (4.0 x 10^3) = 0.5668 x 82.0575 x 265
P= 3.08 atm

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if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The mass of oxygen in the oxide is:

$mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g$

We can calculate the mass percent of O in MgO using the following expression.

$\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\% = 40.0 \%$

You can learn more about mass percent here: brainly.com/question/14990953

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What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?

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237.2 mL.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HCl = (XMV) NaOH.

For HCl; X = 1, M = 0.5 M, V = ??? mL.

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