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snow_lady [41]
2 years ago
6

Rainforests may be located in what latitude range?

Chemistry
1 answer:
SpyIntel [72]2 years ago
6 0

Answer:

near the equator where it's hot

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Soft of easy help plz<br><br><br><br> B<br><br> C<br><br> A<br><br> D
umka21 [38]

Answer:

C

Explanation:

8 0
3 years ago
Which of the following most likely happens when the volume of a gas increases?
mafiozo [28]

the pressure of the gas increases

6 0
2 years ago
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
2 years ago
If an ideal gas has a pressure of 4.03 atm,
kkurt [141]

Answer:

PV=nRT where P=pressure in atm, V=volume is liters, n=numbber of moles, R=gas constant, 0.08206 L-atm/mole KL, and T=temperature in K (273 + C).  So (5.67atm)(99.39L)=n(0.08206 L-atm/mol.K)(328.94K), solve for n, the number of moles, n=20.9 moles.

Explanation:

7 0
2 years ago
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