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3 years ago

7

A compound containing 5.93% H and 94.07% O has a molecular mass of 34.02 g/mol determine the empirical and Molecular formula Of

this compound

1 answer:

Tom [10]3 years ago

7 0

Answer:

7% 4)2(10

Explanation:

beacouse if you divide it you can get the answer

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A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

irakobra [83]

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

<em>0.00500 moles HCl</em>

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>

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2 years ago

In Thomson's plum pudding model the mass of the atom is in the A.electrons. B.matter between the electrons. C.nucleus. D.proton

balu736 [363]

A electrons is the best choice!

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3 years ago

Read 2 more answers

5Br−+BrO3−+6H+→3Br2+3H2O

sashaice [31]

Explanation :

The balanced chemical reaction is,

$5Br^-+BrO_3^-+6H^+\rightarrow 3Br_2+3H_2O$

The expression for the rates of consumption of the reactants are:

The rate of consumption of $Br^-$ = $-\frac{1}{5}\frac{d[Br^-]}{dt}$

The rate of consumption of $BrO_3^-$ = $-\frac{d[BrO_3^-]}{dt}$

The rate of consumption of $H^+$ = $\frac{1}{6}\frac{d[H^+]}{dt}$

The expression for the rates of formation of the products are:

The rate of consumption of $Br_2$ = $+\frac{1}{3}\frac{d[Br_2]}{dt}$

The rate of consumption of $H_2O$ = $+\frac{1}{3}\frac{d[H_2O]}{dt}$

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3 years ago

How much energy must a 10 gram block of ice gain in order to melt ?

Minchanka [31]

Answer:

the answer is 10 times

Explanation:

because it takes 10 times as much energy -3330 j - to melt 10.0 grams of ice.

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2 years ago

How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -><br> 4A1 + 302

Evgesh-ka [11]

<h3>Answer:</h3>

$\displaystyle 40 \ mol \ Al$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 40 \ mol \ Al$

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

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3 years ago