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3 years ago

According to Newton’s first law of morion when will an object at rest begin to move?

Chemistry

1 answer:

Sati [7]3 years ago

6 0

Hello There!

According to "Newton's First Law" an object at rest will begin to move when it is acted upon an unbalanced force.

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Waves crash on a beach one after another. Why doesn't water pile up on the beach

Klio2033 [76]

sand

Explanation:

it is absorbing it

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2 years ago

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2NaCl + F2 yields 2NaF + Cl2

bearhunter [10]

Study your experiment setup. In 30 minutes, how will the air temperature in the bottles compare? What do you predict will happen to the ice in each bottle?

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2 years ago

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$

6 0

2 years ago

What happens when a solid is dissolved in a liquid

slava [35]

the solid particles take up the intermolecular spaces in the liquid.

8 0

3 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago