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Irina18 [472]
3 years ago
5

(03.06 HC)

Chemistry
2 answers:
MrRa [10]3 years ago
3 0

Answer:

higher than 59 c because dipole- dipole interactions in iodine monocholoride are stronger than dispersion forces in bromine

Triss [41]3 years ago
3 0

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

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True are false Selective breeding is a type of genetic engineering.
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Answer:

true

Explanation:

it always an enginering

8 0
3 years ago
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Tính [H] và [OH] của dung dịch HC1 0,1M? Tính pH của dung dịch?
Brrunno [24]

Answer:

po jakiemu to jest XD

Explanation:

po jakiemu to jest XD

3 0
3 years ago
The standard enthalpy change for the following reaction is 873 kJ at 298 K.
Flura [38]

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

2KCl(s)\rightarrow 2K(s)+Cl_2(g)  \Delta H_1=873kJ

Now we have to determine the value of \Delta H for the following reaction i.e,

K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s) \Delta H_2=?

According to the Hess’s law, if we divide the reaction by half then the \Delta H will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=\frac{1}{2}\times (-873kJ)

\Delta H_2=-436.5kJ

Hence, the value of \Delta H_2 for the reaction is -436.5 kJ.

8 0
3 years ago
3 points
vfiekz [6]

Answer:

density=6.74g/ml

:320g÷47.5ml

d=6.74g/ml

thank you

<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>

8 0
3 years ago
A balloon filled with helium occupies 21.0ft³ at 55.0°F. Find the temperature of the helium if the volume of the balloon decreas
harkovskaia [24]

Answer:

Assuming pressure is held constant,the question reduces to a ratio and proportion type of question where;

At constant pressure,

21.0ft³-55.0°F

11.0fy³=11.0ft³/21.0ft³×55°F

The temperature is 28.809°F≈29°F

6 0
3 years ago
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