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zhannawk [14.2K]
2 years ago
3

A chemist wants a 0.100 m solution of naoh. She has 0.0550 mol naoh available. What volume of solution is needed to make the mix

ture?
Chemistry
2 answers:
Veronika [31]2 years ago
5 0

Answer:

She has to mix 0.0550 moles of NaOH in a 550 mL solution to make a 0.100 M solution

Explanation:

Step 1: Data given

Concentration of the solution = 0.100 M

Number of moles NaOH = 0.0550 moles

Step 2: Calculate the volume of solution she needs

Concentration NaOH solution = number of moles NaOH / volume solution

Volume of the solution = Number of moles NaOH / Concentration NaOH solution

Volume of the solution = 0.0550 moles NaOH / 0.100 mol /L

Volume of the solution needed = 0.55 L

Volume of the solution needed = 550 mL

She has to mix 0.0550 moles of NaOH in a 550 mL solution to make a 0.100 M solution

Pavlova-9 [17]2 years ago
5 0

Answer: It's 0.550L.

Explanation: I just did it on edge.

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In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
drek231 [11]

The enthalpy change of the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction is

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To find the enthalpy change we need to apply the following equation

\Delta H =\frac{Q}{n}

To find the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Now, to find the number of moles that react (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

Having these two values we can plug in the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a nice day!

5 0
2 years ago
A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t
leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = P_1=1.0 atm

Initial temperature of the oxygen gas = T_1=20^oC =293.15 K

Final pressure of the oxygen gas = P_2=?

Final temperature of the oxygen gas = T_2=100^oC =373.15 K

Using Gay Lussac's law:

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

P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

P_1V_1=nRT_1 (ideal gas equation)

n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0
2 years ago
Which choice below correctly completes the sentence? _______ reaches the Earth's surface through _______, then turns into ______
blondinia [14]

Answer:

d , before the molten rock becomes lava, it is first magma, and most people know that lava is ejected from volcanoes

4 0
2 years ago
6K + B2O3 → 3K2O + 2B
atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃    →    3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of  B₂O₃ with B from balance chemical equation:

                 B₂O₃          :          B

                    1              :          2

                52.42         :        2×52.42 = 104.84

Thus from 3650 g of  B₂O₃  104.84 moles of boron will produced.

6 0
2 years ago
The three aspects of the measuring process are units, systems, and instruments.
lawyer [7]

Answer:

the answer for that is false

4 0
1 year ago
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