All categories

3 years ago

a balloon has a volume of 1.25 L at 1.5 atm of pressure. What is the volume on the baloon and 0.25 atm of pressure?

Chemistry

1 answer:

kvv77 [185]3 years ago

5 0

Answer: 7.5 L

Explanation: According to Boyle's law, V1P1=V2P2. This means that

(1.25)(1.5)=(V2)(.25), so 1.875/.25=V2, so the volume of the baloon at .25 atm is equal to 7.5 L.

You might be interested in

How many moles of CO2 are produced from the combustion of 6.40 mol c2h8?

jarptica [38.1K]

12.8 mole of CO2 from the combustion

7 0

2 years ago

What role do hydrogen ions play in the electron transport chain high?

jeka94

They are pumped across the mitochondrial inner membrane against their concentration gradient (to where their concentration is high); as the H+ ions flow back to where their concentration is low, they drive ATP synthase to form ATP

7 0

3 years ago

g How many turns through the citric acid cycle are required to fully oxidize all of the acetly-coA that result from 1 molecule o

faust18 [17]

Answer:

The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.

Explanation:

Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:

Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺

Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.

2 Pyruvate ----> 2 AcetylCoA + 2CO₂

Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.

7 0

3 years ago

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

tensa zangetsu [6.8K]

<h3>Answers:</h3>

1) 2 Units of Ozone

2) 3 Units of Ozone

3) 9 Units of Ozone

<h3>Solution:</h3>

1) From 6 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

6 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

9 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 3 Units of Ozone

3) From 27 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

27 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 9 Units of Ozone

3 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago