Assuming a 100% dissociation, what is the freezing point and boiling point of 2.59 m K3PO4(aq).

Answers

Answer 1

Answer: Assuming 100% dissociation of K3PO4 you get each molecule of K3 PO4 dissociates into four particles: 3 K4 cations and 1 PO4 - anion. That means that Van't Hoff factor, i, is 4. i = 4. So, the decrease in freezing point of the water solution is Delta Tf = i * Kf * m. Where Kf is the molal cryoscopic constant of water = 1.86 °C /m; and the molality m = 2.59 m => Delta Tf = 4 * 1.86 °C / m * 2.59 m = 19.3 °C. And the freezing point is the normal freezing points less 19.3°C = 0°C - 19.3°C = - 19.3°C. The increase in the boiling point Delta Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.

Answer 2

Answer:

Answer: Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.

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Leutium-176 has a half-life of 3.85 mc012-1.jpg 1010 years. After 1.155 mc012-2.jpg 1011 years, how much leutium-176 will remain from an original 16.8-g sample?2.10 g
3.00 g
5.56 g
8.40 g

Answers

Answer : The amount left of leutium-176 will be, 2.10 g

Solution :

First we have to calculate the rate constant, we use the formula :

$k=(0.693)/(t_(1/2))$

$k=\frac{0.693}{3.85* 10^(10)\text{years}}$

$k=0.18* 10^(-10)\text{years}^(-1)$

Now we have to calculate the amount left of the sample.

Expression for rate law for first order kinetics is given by :

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant = $0.18* 10^(-10)\text{years}^(-1)$

t = decay time = $1.155* 10^(11)\text{ years}$

a = initial amount of the sample = 16.8 g

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$1.155* 10^(11)\text{years}=\frac{2.303}{0.18* 10^(-10)\text{years}^(-1)}\log(16.8)/(a-x)$

$a-x=2.10g$

Therefore, the amount left of leutium-176 will be, 2.10 g

Answer:

Explanation:

Given the balanced equation representing a reaction:2C2H6 + 7O2==>4CO2 + 6H2O

Determine the total number of moles of oxygen that react completely with 8.0 moles of C2H6.

Answers

The number of moles of O₂ will be 28 moles.

The given balanced reaction:

$2C_2H_6 + 7O_2==>4CO_2 + 6H_2O$

From the above reaction,

2 moles of C₂H₆ nd 7 moles of O₂ reacts together to produce 4 moles of CO₂ and 6 moles of H₂O.

Given:

Number of moles of C₂H₆ = 8

To find:

Number of moles of O₂=?

2 moles of C₂H₆ reacts with 7 moles of O₂.

1 mol of C₂H₆ = $(7)/(2)$ moles of O₂

8 moles of C₂H₆ = $(7)/(2) *8$ moles of O₂

Thus, number of moles of O₂ will be 28 moles.

Find more information about "Number of moles" here:

brainly.com/question/13314627

Explain how a hawk or eagle could use convection currents to soar upward without flapping its wings

Answers

Hawks will usually fly in a particular area and circle it many times right?Therefore these birds gain their altitude by riding in the current air with other surrounding masses,thats why birds can go high up without using too much energy

What can the arrow in a chemical reaction be translated to mean?

Answers

When a chemical equation is written out, there will be an arrow to represent the chemical reaction occurring. In the equation 6H20+6CO2-->6C12H6O+6O2, the part before the arrow is what you start out with. Next comes the arrow, or the reaction. After the reaction, you have the end product.

Answer:

The arrow means "yield" or "react to form".

Explanation: The arrow means "yield" or "react to form".

Which element, represented by X, reacts with fluorine to produce the compound XF2?(1) aluminum (3) magnesium
(2) argon (4) sodium

Answers

The answer is (3) magnesium. Aluminum usually has valence of +3 and Sodium usually has valence of +1. Argon is usually not react with others. Because it is already stable with 8 electrons at the outer shell.

Science fair ideas???

Answers

You can use a website called sciencebuddies to find a project

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If a water faucet drips 15.0 times per minute, how many gallons will pass into the drain in 2.25 years. Assume each drop = 0.20 mL