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Alisiya [41]
2 years ago
8

How many molecules of PF5 are found in 39.5 grams of PF5?

Chemistry
1 answer:
djyliett [7]2 years ago
7 0

Answer:

1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

Explanation:

Atomic weights : P= 31, F= 19,

molar mass of = 1 atomic weight of P+ 5 atomic weight of

 F= 31+5 \times 19

= 31+95

 =126 g/mole

moles in 39.5 gm of

= \frac{Mass}{Molar mass}

 = \frac{39.5}{126}moles

 =0.3134 moles

1 mole of any substance contains

0.3131 moles contains 0.3134  

 = 1.9 \times 10^{23} molecules

Therefore, 1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

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If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen
Nikolay [14]

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

2NOBr(g)\rightarrow 2NO(g)+Br_2(g)

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.0440 M

Putting values in above equation, we get:

0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)

[A]=0.00734M

Hence, the concentration of after 9.0 seconds is, 0.00734 M

5 0
2 years ago
When 200 grams of water cools from 50.°C to 25°C, the total amount of heat energy released by the water is?
Ierofanga [76]
We will use the formula q = mCΔT. where q = heat energy released. m= mass. C= specific heat of water. ΔT= change in temperature. Specific heat of water has not been given so we will just use the standard value of that which is 4.186J/g °C. q = mCΔT. q = 200 × 4.186 × (50 -25) q = 200 × 4.186 × 25 q = 5000 × 4.186 q = 20, 930 The total amount of energy released is 20, 930 J
6 0
2 years ago
Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv
faltersainse [42]

Answer:

Solution of isopropanol is 10.25 molal

Explanation:

615 g of isopropanol (C3H7OH) per liter

We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

Density of water 1g/mL → Density = Mass of water / 1000 mL of water

Notice we converted the L to mL

Mass of water = 1000 g (which is the same to say 1kg)

Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol  → 615 g . 1mol / 60g = 10.25 moles

Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m

4 0
2 years ago
If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc
igor_vitrenko [27]
When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions. 
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>

7 0
2 years ago
an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the
ziro4ka [17]

Answer:

The specific heat capacity of the metal is 0.843J/g°C

Explanation:

Hello,

To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.

Heat gained by the metal = heat loss by water + calorimeter

Data,

Mass of metal (M1) = 512g

Mass of water (M2) = 325g

Initial temperature of the metal (T1) = 15°C

Initial temperature of water (T2) = 98°C

Final temperature of the mixture (T3) = 78°C

Specific heat capacity of metal (C1) = ?

Specific heat capacity of water (C2) = 4.184J/g°C

Heat loss = heat gain

M2C2(T2 - T3) = M1C1(T3 - T1)

325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)

1359.8 × 20 = 512C1 × 63

27196 = 32256C1

C1 = 27196 / 32256

C1 = 0.843J/g°C

The specific heat capacity of the metal is 0.843J/g°C

8 0
2 years ago
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