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3 years ago

7

When taking liquids from a reagent bottle

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

B. pour the reagent into a small beaker before using it

Explanation:

Let's explore all the mentioned practice

A. Leftover reagents should not be discarded in the sink but discarded according to the disposal method present in the safety data sheet (SDS) of the chemical.

B. Reagents should be poured in a clean beaker for use to avoid contamination in the original reagent bottle

C. This is not a good practice, as the plastic pipet tends to contain some impurity in its tip.

D. This is also a bad practice. Firstly, a reagent should be taken according to the need. If there is leftover, it should be discarded as it most probably contains contamination in it

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A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v

Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

$0.9762bar.\frac{1atm}{1.013bar} =0.9637atm$

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol$

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:

$\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol$

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4 years ago

You need a 30% alcohol solution. On hand, you have a 200 mL of a 10% alcohol mixture. You also have 55% alcohol mixture. How muc

Art [367]

Answer: 160 ml

Explanation:

The expression used will be :

$C_1V_1+C_2V_2=C_3V_3$

where,

$C_1$ = concentration of Ist alcohol solution= 10%

$C_2$ = concentration of 2nd alcohol solution= 55%

$V_1$ = volume of Ist alcohol solution = 200 ml

$V_2$ = volume of 2nd alcohol solution= v ml

$C_3$ = concentration of resulting alcohol solution= 30%

$V_2$ = volume of resulting alcohol solution= (v+200) ml

Now put all the given values in the above law, we get the volume of added.

$(10\times 200)+(55\times v)=(30\times (v+200)ml)$

By solving the terms, we get :

$v=160ml$

Therefore, the volume of 55% mixture needed to be added to obtain the desired solution is 160 ml.

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3 years ago

EASY BUT I STILL NEED HELP

Jet001 [13]

Answer:

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Explanation:

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Alexxandr [17]

Either a gas or a vapor

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Anastaziya [24]

Answer:

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