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Mekhanik [1.2K]
3 years ago
13

How should students prepare to use chemicals in the lab? Select one or more: Sort the lab chemicals in alphabetical order for qu

ick access. Become familiar with the chemicals to be used, including exposure or spill hazards. Locate the spill kits and understand how they are used. Check labels and discard chemicals that are not 100% pure. CHECK
Chemistry
1 answer:
maksim [4K]3 years ago
7 0

Answer:

Sort the lab chemicals in alphabetical order for quick access.

Become familiar with the chemicals to be used, including exposure or spill hazards.

Locate the spill kits and understand how they are used.

Explanation:

There are many chemicals in a laboratory hence they should be sorted out and arranged in alphabetical order so that theory can easily be identified and located whenever they are required.

The properties of each chemical should be known especially hazards connected to exposure or spill of the chemicals.

The students should also familiarize themselves with the contents of spill kits and how they are used.

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20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

To know more about Percentage Yield

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5 0
1 year ago
Which element has the greatest electronegativity?
Alex Ar [27]
Fluorine has the highest. Fluorine's electronegativity is 4.0
3 0
3 years ago
I need to solve for x in this equation<br><br> ln(760/630)=32/8.314(1/x-1/329.5)
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4 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
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