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2 years ago

Suggest why sodium and hydrogen ions do not diffuse at the same rate

Chemistry

1 answer:

Troyanec [42]2 years ago

8 0

Answer:

sodium has got ionic bonds that are weak

compared to hydrogen covalent bonds that are strong

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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:

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

Thus the enthalpy change for the given reaction is -705.55kJ

7 0

2 years ago

In a reverse reaction, products are formed from reactants and reactants are formed from products. when they yield the same amoun

GREYUIT [131]

.......When they yield the same amount , *Reversible reactions* will reach chemical equilibrium........

8 0

2 years ago

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You heat 51 grams of magnesium over a Bunsen burner for several minutes until it reacts with oxygen in the air. Then you weigh t

garik1379 [7]

Answer:

The mass was there all along, it was just in the air. The weight of the oxygen from the air is not weighed in the beginning, only at the end as part of the product, making it seem like there is a total mass change.

8 0

2 years ago

Estimate the following: a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. b) The mass of ethylene contained in a 0

Nina [5.8K]

Explanation:

(a) The given data is as follows.

T = $55^{o}C$ = (55 + 273) K

= 328 K

P = 35 bar = 3500 kPa

Let us assume that moles of ethylene present are 1 kmol and according to the ideal gas equation, PV = nRT.

Or, V = $\frac{nRT}{P}$

= $\frac{1 \times 8.314 \times 328}{3500 kPa}$

= 0.779 $m^{3}$

Hence, the volume occupied by 18 kg of ethylene at $55^{o}C$ and 35 bar is 0.779 $m^{3}$ .

(b) The given data is as follows.

V = 0.25 $m^{3}$

T = $50^{o}C$ = (50 + 273) K

= 323 K

P = 115 bar = 11500 kPa

Using ideal gas equation first, we will calculate its moles as follows.

n = $\frac{PV}{RT}$

= $\frac{11500 \times 0.25}{8.314 \times 323 K}$

= 1.07 mol

Since, moles = $\frac{mass}{\text{molar mass}}$

Hence, mass of ethylene will be calculated as follows.

moles = $\frac{mass}{\text{molar mass}}$

1.07 mol = $\frac{mass}{28 g/mol}$

mass = 29.96 g

Therefore, mass of given ethylene is 29.96 g.

8 0

2 years ago

Fermium-253 is a radioactive isotope of fermium that has a half-life of 3.0 days. A scientist obtained a sample that contained 2

Dahasolnce [82]

Problem 2

You start out with 216 ugrams of Fermium - 253. After 3 days, you will have 1/2 as much. 108 ugrams is what you have.

Another 3 days goes by. You started with 108 ugrams. That gets cut in 1/2 again. Now you have 54 ugrams.

Finally another 3 days goes by. You started with 54 ugrams. you now have 1/2 as much which would be 27 ugrams

#days Amount in micrograms

0 216

3 108

6 54

9 27

Problem One

You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.

Table

Bond Energy Kj/Mol Bond Length pico meters

N - N 167 145

N=N 418 125

N≡N 942 110

Rules

As the number of bonds INCREASES, the energy contained in the bond goes UP

As the number of bonds INCREASES, the length of the bond goes DOWN.

5 0

2 years ago

Read 2 more answers