The temperature will decrease the rate of reactions.

How many moles are in 4.818 x 1024 chloride ions

erastovalidia [21]

Answer:

<h3>The answer is 8.00 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{4.818 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 8.00332225...$

We have the final answer as

<h3>8.00 moles</h3>

Hope this helps you

4 0

3 years ago

How many grams of water would require 2200 joules of heat to raise it’s temperature from 34 degrees Celsius to 100 degrees celsi

Arada [10]

Answer:

Mass, m = 7.97 grams

Explanation:

We have,

Initial temperature was 34 degrees Celsius

Heat produced in it is 100 calories of heat

The specific heat capacity of water is 4.18 J/g °C

The heat produced due to change in temperature is given by :

$Q=mc(T_f-T_i)$

m is mass of water in grams

$m=\dfrac{Q}{c(T_f-T_i)}\\\\m=\dfrac{2200}{4.18\times (100-34)}\\\\m=7.97\ g$

So, 7.97 grams of water is required to produce heat of 2200 joules.

7 0

3 years ago

What volume in milliliters of 0.0130 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl?

Orlov [11]

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That girl, is a gun to my holster
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8 0

3 years ago

What would be the freezing point of a solution that has a molality of 1.468 m which was prepared by dissolving biphenyl (C12H10)

Readme [11.4K]

Answer:

Freezing point solution = 70.131 °C

Explanation:

Step 1: Data given

Molality = 1.468 molal

A solution is created by dissolving biphenyl (C12H10) into naphthalene

Biphenyl is a non-electrolyte

Freezing point of naphthalene = 80.26 °C

Step 2: Calculate the freezing point depression

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of biphenyl = 1

⇒with Kf = the freezing point depression constant of naphthalene = 6.90 °C/m

⇒with m = the molality = 1.468 molal

ΔT = 1 * 6.90 °C/m * 1.468 °C

ΔT = 10.13 °C

Step 3: Calculate the freezing point of the solution

ΔT = 10.13 °C

Freezing point solution = freezing point naphthalene - 10.13 °C

Freezing point solution = 80.26 °C - 10.129 °C

Freezing point solution = 70.131 °C

5 0

3 years ago

What is the total number of molecules in 11.2 liters of N2 gas at STP

scoray [572]

Molar volume at STP = 22,4 L

1 mole -------------- 22,4 L
x mole -------------- 11,2 L

x = 11,2 / 22,4

x = 0,5 moles of N2

1 mole --------------- 6,02.10²³ molecules
0,5 moles ------------ y molecules

y = 0,5 . 6,02.10²³

y = 3,01.10²³ molecules

7 0

3 years ago