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Digiron [165]
4 years ago
7

What bonds break? select that apply

Chemistry
1 answer:
igor_vitrenko [27]4 years ago
7 0
Hydrogen-Hydrogen
Oxygen=Oxygen
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Consider the following intermediate chemical equations.
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Answer: 250 kJ

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According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ   (4)

8 0
3 years ago
Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s​
vfiekz [6]

To Find :

The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s​.

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The horizontal range of a projectile is given by :

R = \dfrac{u^2 sin 2\theta}{g} ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m

Therefore, the horizontal range of projectile is 80 m.

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All of the following are examples of minerals except __________.
snow_tiger [21]
The answer is D.timber good luck 

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