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konstantin123 [22]
2 years ago
11

you are carrying out the following reaction: o2 2h2 −−> 2h2o you start with 6.0 moles of oxygen and 4.0 moles of hydrogen. ho

w many moles of water will you make?
Chemistry
1 answer:
Vinil7 [7]2 years ago
4 0
Chemistry is speak about how world goin in
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What is common to all elements in the same group on the periodic table
marin [14]
They all have the same number of electrons.
3 0
2 years ago
Read 2 more answers
A student fails to clean the pipet first. after delivering the vinegar sample, the student notices a drop of vinegar clinging to
VladimirAG [237]
This should not matter because the pipet has gradations and usually more of the sample is taken up in the pipette than what is delivered into the flask the student should always rinse the container being used because they are contaminating the sample if they do not clean it out
4 0
2 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
2 years ago
O tetraclorometano é um solvente muito utilizado na indústria de transformaç?o de produtos orgânicos ele é produzido pela reaç?o
Vlada [557]

Answer:

ΔHr = -103,4 kcal/mol

Explanation:

<u>Using:</u>

<u>AH° (kcal/mol) </u>

<u>Metano (CH) </u>

<u>-17,9 </u>

<u>Cloro (CI) </u>

<u>tetraclorometano (CCI) </u>

<u>- 33,3 </u>

<u>Acido cloridrico (HCI) </u>

<u>-22</u>

It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:

ΔHr = (ΔH products - ΔH reactants)

For the reaction:

CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)

The balanced reaction is:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

The ΔH's of formation for these compounds are:

ΔH CH₄(g): -17,9 kcal/mol

ΔH Cl₂(g): 0 kcal/mol

ΔH CCl₄(g): -33,3 kcal/mol

ΔH HCl(g): -22 kcal/mol

The ΔHr is:

-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)

<em>ΔHr = -103,4 kcal/mol</em>

<em></em>

I hope it helps!

3 0
2 years ago
Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
2 years ago
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