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3 years ago

What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2 ? ...?

1 answer:

8 0

To make a first step you have to know the balanced form for neutralization formula: $Mg(OH)2(aq)+2HCl(aq)--\ \textgreater \ 2H2O+Mg(aq)+Cl(aq)
$
According to this, you can <span>calculate what you are being asked :</span> $0.0442molMg(OH)2(x)(2molHCl)/(1molMg(OH)2)=0.0885molHCl$
Then we have : $0.140MHCl=0.140(mol/L)HCl$
Hope everything is clear, here is the exact answer you need : $V=(0.0885molHCl)/(0.140(mol/L)HCl)=0.632LHCl
$

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How many covalent bonds are in a nitrogen molecule?

makvit [3.9K]

3 covalent bonds (there are 2 electrons in the first orbital and 5 in the second. You still have room for three more)

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3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

A 13.00 g sample of citric acid reacts with an excess of baking soda as shown in the equation.

PolarNik [594]

Answer:

eheheehehehszndn!jejxxnndrrjrrrfufurururufjththjrjrjdjjjrj\u\ujrjeejrjjjj carbon

4 0

2 years ago

Read 2 more answers

Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting mate

Troyanec [42]

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

8 0

3 years ago

Read 2 more answers

An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?

aleksley [76]

Answer:- Formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

$81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})$

= $0.204molFe_2(SO_4)_3$

$18.4gH_2O(\frac{1mol}{18.02g})$

= $1.02molH_2O$

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

$Fe_2(SO_4)_3=\frac{0.204}{0.204}$ = 1

$H_2O=\frac{1.02}{0.204}$ = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and the name of the hydrate is Iron(III)sulfate pentahydrate.

3 0

3 years ago