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2 years ago

15.35 L of gas is at a pressure of 198.0 kpa. At what pressure would the gas have a volume of 13.78

Chemistry

1 answer:

Ivahew [28]2 years ago

5 0

Answer:

P₂ = 220.56KPa

Explanation:

Boyles Law P ∝ 1/V => Inverse relationship => Decreasing Volume => Increasing Pressure.

P₁ = 198 KPa P₂ = ?

V₁ = 15.35 Liters V₂ = 13.78 Liters

P₁V₁ = P₂V₂ => P₂ = P₁(V₁/V₂) = 198KPa(15.35 L/ 13.78L) = 220.6KPa

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Suppose you had a balloon containing 1 mole of helium at STP and a balloon containing 1 mole of oxygen at STP. Which statement(s

AleksAgata [21]

Answer:

The true statement is option A.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 1 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K

$V=\frac{nRT}{P}=\frac{1 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}$

V = 22.42 L

This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.

So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.

7 0

2 years ago

in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally

Gwar [14]

Answer:

The average atomic mass of bromine is 79.9 amu.

Explanation:

Given data:

Percentage of Br⁷⁹ = 55%

Percentage of Br⁸¹ = 45%

Average atomic mass of bromine = ?

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [55 × 79] + [81 ×45] / 100

Average atomic mass = 4345 + 3645 / 100

Average atomic mass = 7990 / 100

Average atomic mass = 79.9 amu

The average atomic mass of bromine is 79.9 amu.

3 0

2 years ago

What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi

kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

$\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s)$ .

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

$\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq)$ .

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

How to rewrite a chemical equation to produce a net ionic equation?

Rewrite all reactants and products that ionizes completely in the solution as ions.
Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

Soluble salts,
Strong acids, and
Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as $\rm Mg^{2+}$ and $\rm 2\; Cl^{-}$ .
NaOH is a strong base. Each unit of NaOH can be written as $\rm Na^{+}$ and $\rm OH^{-}$ .
Mg(OH)₂ is a weak base and should not be written.
NaCl is a salt and is soluble in water. Each unit of NaCl can be written as $\rm Na^{+}$ and $\rm Cl^{-}$ .

$\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}$ .

Ions on both sides of the equation:

$\rm 2\; Cl^{-}$ , and
$\rm 2\; Na^{+}$ .

Add the state symbols:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

For the second reaction:

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

$\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}$ .

$\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s)$ .

5 0

2 years ago

Read 2 more answers

Draw the structures of organic compounds A and B. Indicate stereochemistry where applicable The starting material is ethyne, a c

k0ka [10]

Answer:

See explanation below

Explanation:

In this case we have the starting reactant which is the ethine, In the first step reacts with NaNH₂, a strong base. This base will substract the hydrogen from one of the carbon of the ethine, and form a carbanion. This will react with the propane bromide, displacing the bromine and forming a 5 carbon chain with the triple bond on the carbon 1 and 2.

In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.

Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.

The whole process can be seen in the picture below.

Hope this helps

3 0

2 years ago

How much energy is required to melt a 20.0lb bag of ice a 0°c? A pound (lb.) is equivalent to 0.4536. The Hfusion of ice is +6.0

erma4kov [3.2K]

To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have

$Q = n \Delta H_{f}$

where the latent heat of fusion for ice is equal to 6.009 kJ/mol.

Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.

To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have

$ice_{moles} = (\frac{9072 g}{1})(\frac{1 mol}{18.01 g}) =503.72 moles$

Now, to compute for the molar heat of fusion, Q,

$Q = (503.72)(6.009) = 3026.9 kJ$

Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.

Answer: 3026.9 kJ

6 0

2 years ago