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3 years ago

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Step I of the proposed mechanism involves the collision between NO2 and F2 molecules. This step is slow even though such collisi

ons occur very frequently in a mixture of NO2(g) and F2(g). Consider a specific collision between a molecule of NO2 and a molecule of F2.

1 answer:

pogonyaev3 years ago

7 0

Answer:

bob

Explanation:

bob

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What is the number of copper atom in one naira coin which weight is 7.3g a sum of the material is 86%

Step2247 [10]

Answer:

5.9x10²² atoms of Cu are in one naira coin

Explanation:

To solve this question we need to find the mass of Copper. Then, using its molar mass (Cu = 63.546g/mol) we must find the moles of Cu and its atoms using Avogadro's number:

<em>Mass Cu:</em>

7.3g * 86% = 6.278g is the mass of Cu.

<em>Moles Cu:</em>

6.278g * (1mol / 63.546g) = 0.099moles Cu

<em>Atoms Cu:</em>

0.099moles Cu * (6.022x10²³atoms / 1mol) =

<h3>5.9x10²² atoms of Cu are in one naira coin</h3>

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2 years ago

The process in which water changes from a liquid to a vapor is known as __________.

Basile [38]

Hello!
The answer is Evaporation!

When you add enough heat to a liquid, it boils turning into a gas. This is called Evaporation

7 0

3 years ago

Read 2 more answers

How many moles of N2 in 57.1 g of N2?

SpyIntel [72]

We are given –

Mass of $\bf N_2$ is 57.1 g and we are asked to find number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\pink{\bf\longrightarrow { Molar \:mass \:of \: N_2:-} }$

$\qquad$ $\bf \twoheadrightarrow 14\times 2$

$\qquad$ $\bf \twoheadrightarrow 28$

$\qquad$ ____________________

Now,Let's calculate the number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\purple{\bf\longrightarrow { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}$

$\qquad$ $\bf \twoheadrightarrow \dfrac{57.1}{28}$

$\qquad$ $\bf \twoheadrightarrow 2.04\: moles$

__________________________________

7 0

2 years ago

Consider the chemical equations shown here.

vichka [17]

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Explanation:

The overall equation for the reaction that produces P₄0₁₀ is :

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Now let us derive this equation:

Given equations:

P₄ $_{s}$ + 30₂ $_{g}$ ⇒ P₄0₆ $_{s}$ equation 1;

P₄ $_{s}$ + 50₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$ equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

P₄0₆ $_{s}$ ⇒ P₄ $_{s}$ + 30₂ $_{g}$ equation 3

+

equation 2:

P₄ $_{s}$ + 50₂ $_{g}$ + P₄0₆ $_{s}$ ⇒ P₄0₁₀ $_{s}$ + P₄ $_{s}$ + 30₂ $_{g}$

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

5 0

2 years ago

How many molecules of CO are produced when 2.73 moles of HPO3 react? 4HPO3 + 12C ----> 2H2 + 12CO + P4

Zina [86]

According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃ ---gives---> 12 moles of CO
2.73 moles of HPO₃ ---gives---> ? moles of CO
so number of moles of CO = $\frac{(2.73 * 12)}{4}$ = 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
= 8.19 * (6.022 * 10²³) = 4.93 * 10²⁴ molecules

3 0

3 years ago