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Vera_Pavlovna [14]
1 year ago
9

Isotopes of an element differ due to the number of.

Chemistry
1 answer:
Alchen [17]1 year ago
7 0

Answer:

Neutrons

Explanation:

Isotopes are variations of an element that differ from the number of neutrons.

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3 years ago
Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

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7 0
3 years ago
The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to
sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

2

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

c

=

n

V

So, how many moles of magnesium chloride must be present in the target solution?

c

=

n

V

⇒

n

=

c

⋅

V

n

=

0.158 M

⋅

250.0

⋅

10

−

3

L

=

0.0395 moles MgCl

2

Now determine what volume of the target solution would contain this many moles of magnesium chloride

c

=

n

V

⇒

V

=

n

c

V

=

0.0395

moles

3.15

moles

L

=

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

V

=

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

.

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sample of the stock solution by a dilution factor of

20

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