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Yuri [45]
2 years ago
13

For an experiment, a student needs 100.0 mL of 0.4220 M NaCl. If the student starts with NaCl(s) and distilled water, which of t

he following pieces of laboratory glassware should the student use to prepare the solution with the greatest accuracy?A) 25 mL volumetric pipetB) 100 mL erienmeyer flask C) 100 mL graduated cylinder D) 100 mL volumetric flask E) 1 L beaker
Chemistry
1 answer:
pochemuha2 years ago
6 0

Answer:

100 mL volumetric flask.

Explanation:

For preparation of any solution we generally use volumetric flask.

We will weigh the required amount of NaCl (based on calculation for 100 mL of 0.422 M NaCl) and will add this in a 100 mL volumetric flask.

There is only one mark in the volumetric flask.

So we will start adding water to it through wash bottle.

Then will continue stirring the solution to dissolve NaCl with addition of more amount of water (use funnel to add water).

Then finally we will make up the solution to 100 mL of mark.

You might be interested in
A sample that weighs 103.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the samp
aalyn [17]
1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u

Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.

1) We can find  the next equation:
mass=molar mass  x number of moles.

x=number of moles of helium
y=number of moles of helium. 

(4 g/mol) x  +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:

(1)

4x+83.8y=103.75


2) We can find other equation:

We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.

1 mol  is always 6.022 * 10²³ atoms or molecules, (in this case atoms).

Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.

x=30% of (x+y)

Therefore, we have this other equation:
(2)

x=0.3(x+y)


With the equations(1) and (2), we have the next system of equations:

4x+83.8y=103.75

x=0.3(x+y)  ⇒ x=0.3x+0.3y  ⇒    x-0.3x=0.3y  ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7

We solve this system of equations by substitution method.
x=3y/7

4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21

x=3y/7=3(1.21)/7=0.52

We have 0.52 moles of  helium and 1.21 moles of Krypton.

1 mol=6.022 * 10²³ atoms

Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).

Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.

Answer: The sample have 1.044 * 10²⁴ atoms.
6 0
2 years ago
Read 2 more answers
An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
1 year ago
Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo
sleet_krkn [62]

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

5 0
1 year ago
The pressure and temperature inside a bike tire is 10 atm and 10k respectively. What will the pressure become in the tire when t
igomit [66]

Answer:

P_2=20atm

Explanation:

Hello,

In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:

\frac{P_1}{T_1}= \frac{P_2}{T_2}

Thus, we solve for the final pressure P2 to obtain it as shown below:

P_2=\frac{P_1T_2}{T_1}=\frac{10atm*20K}{10K} \\\\P_2=20atm

Hence, we notice that the temperature doubles as well as the pressure.

Best regards.

3 0
2 years ago
Part a write an equation for the formation of nh3(g) from its elements in its standard states. express your answer as a chemical
andrey2020 [161]
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
3 0
2 years ago
Read 2 more answers
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