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3 years ago

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Help me real quick !!

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

yellow

if not sorry that's though the best answer

Explanation:

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What is climate? How does it differ from weather

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Weather refers to short term atmospheric conditions while climate is the weather of a specific region averaged over a long period of time. Climate change refers to long-term changes.

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3 years ago

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How are the molecules moving in a cold glass of water compared to in a warm glass of water

taurus [48]

Molecules move faster in a warm glass of water they move slower in a cold glass of water

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3 years ago

Natasha_Volkova [10]

Answer:

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Explanation:

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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

3 years ago

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

$N= N_0(\frac{1}{2})^n$

where

$n = \frac{t}{t_{\frac{1}{2}}}$

here

t = 458,000 years

$t_{\frac{1}{2}}$ = 229,000

$\frac{t}{t_{\frac{1}{2}}}$ = \ $\frac{ 458,000}{229,000}$

n = $\frac{t}{t_{\frac{1}{2}}}$ = 2.000

Now substituting the values

$1.675 = N_0(\frac{1}{2})^{2.000}}$

$1.675 = N_0\times (0.2500)$

$N_0= \frac{1.675}{0.2500}$

$N_0=6.70$

3 0

3 years ago