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3 years ago

Some elements are solids at room temperature true or false

Chemistry

1 answer:

nadya68 [22]3 years ago

7 0

This is true, think of silver, gold, etc— those are heated to make them liquid

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Which term best describes the stored energy based on the position or chemical composition of an object?

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Stored energy is described as potential energy

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3 years ago

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If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?

SCORPION-xisa [38]

Answer:

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)

As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

Density = Mass / Volume → Density . Volume = Mass

0.00183 g/mL . 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide

Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles

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3 years ago

Choose the number of significant figures indicated. 0.078

MAXImum [283]

2 significant figures

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3 years ago

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Why is there no set path that a scientific inquiry must follow

Marina CMI [18]

Due to the variables that each inquiry has, Scientific inquiry follows a path of questioning and testing a hypothesis, however this changes in response to specific details.

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2 years ago

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The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)

Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

$\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)$

B. The rate expressions

$-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]$

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

$\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}$

D. Substitute this concentration into the rate law

$\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]$

The reaction is second order in NO and first order in O₂.

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3 years ago