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liq [111]
3 years ago
7

You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How

much solid aluminum sulfate should you add
Chemistry
1 answer:
Ksju [112]3 years ago
8 0

Answer:

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask  

The molar mass of Aluminum Sulfate = 342.15 g/mol  

Using the molarity formula:-  

Molarity = Number of moles/Volume of solution in a liter  

Number of moles = Given weight/ molar mass  

Molarity = (Given weight/ molar mass)/Volume of solution in liter  

0.187 M = (Given weight/342.15 g/mol)/0.250 L  

Given weight = 15.99 g  

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b)Elemental boron (as a gas) is produced in one industrial process by heating diboron trioxide with magnesium metal, also produc
dlinn [17]

<u>Answer:</u> The unbalanced chemical equation is written below.

<u>Explanation:</u>

Unbalanced chemical equation does not follow law of conservation of mass.

In an unbalanced chemical equation, total number of individual atoms on the reactant side will not be equal to the total number of individual atoms on the product side.

The chemical equation for the reaction of diboron trioxide and magnesium metal follows:

B_2O_3(s)+Mg(s)\rightarrow B(g)+MgO(s)

Hence, the unbalanced chemical equation is written above.

5 0
3 years ago
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0
3 years ago
Question 21 of 25
RideAnS [48]

Answer:

7

Explanation:

7 0
3 years ago
Comare and contrast physical property and chemical property
Nimfa-mama [501]
A physical property is a quality or condition of a substance that can be observed or measured without changing the substance's composition. a chemical property is the ability of a substance to undergo a specific chemical change.
8 0
3 years ago
How many grams are there in 1.1 x 1027 molecules of water (H2O)?
Nezavi [6.7K]

Answer:

C) 3.3 x 104 grams

Explanation:

1 mole of water contains 6.02 × 10^23 atoms

1.1 × 10^27 atoms will contain;

1.1 × 10^27 ÷ 6.02 × 10^23

= 0.1827 × 10^( 27 - 23)

= 0.1827 × 10^(4)

= 1.827 × 10³ moles of water.

To convert mole to mass in grams, we use the formula;

mole (n) = mass (m) ÷ molar mass (MM)

Molar mas of water (H2O) = 1(2) of H + 16 of O = 18g/mol

mole = mass/molar mass

1.827 × 10³ = mass / 18

mass = 1.827 × 10³ × 18

mass = 32.886 × 10³

mass = 3.286 × 10⁴

mass = 3.3 × 10⁴ grams

6 0
3 years ago
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