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antiseptic1488 [7]
3 years ago
14

How many grams of sodium chloride are required to make 2.00 L of a solution with a concentration of 0.100 M?

Chemistry
1 answer:
vampirchik [111]3 years ago
4 0

Answer:

Mass = 11.688g

Explanation:

Volume = 2.00L

Molar concentration = 0.100M

Mass = ?

These quantities are relatted by the following equation;

Conc = Number of moles / volume

Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 0.2mol * 58.44g/mol

Mass = 11.688g

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A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?
Pachacha [2.7K]
<span>The student should follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>

</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L

Molarity = number of moles / volume of the solution

Hence, number of moles in 1 L = 2 mol

2. Find out the mass of dry CaCl</span>₂ in 2 moles.<span>

moles = mass / molar mass

Moles of CaCl₂ = 2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol

Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
                                     = 221.96 g

3. Weigh the mass accurately 

4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and finally wash the funnel and watch glass with de-ionized water. That water also should be added into the volumetric flask.

5. Then add some de-ionized water into the volumetric flask and swirl well until all salt are dissolved.

<span>6. Then top up to mark of the volumetric flask carefully. 
</span>
</span>
7. As the final step prepared solution should be labelled.
4 0
3 years ago
Wht is this called<br>N which chain does it belong​
german

Answer:

national

Explanation:

I don't understand, I'm sorry

5 0
3 years ago
An atom of sodium-23 (Na-23) has a net charge of +1. Identify the number of protons, neutrons, and electrons in the atom. Then,
kolezko [41]

Answer:

Number of proton is 11

Number of neutrons is 12

Number of electrons is 10

Explanation:

For a neutral Na, Sodium atom:

The mass number P+N = 23

Atomic number(E or P) = 11

A charged Na atom that has lost an electron is positively charged and so:

Number of proton = 11 (Still the same)

Number of neutrons = 12

Number of electrons = 11-1=10 (the atom has lost an electron)

If an atom loses electron, it becomes positively charged ie P>E

7 0
3 years ago
Read 2 more answers
How much heat (in kilojoules) is evolved or absorbed in the reaction of 1.90g of Na with H2O ?
irina1246 [14]
The heat of the reaction is an extensive property: it is proportional to the quantity of the quantity that reacts.

The change in enthalpy is a measured of the heat evolved of absorbed.

When the heat is released, the change in enthalpy is negative.

The reaction of 2 moles of Na develops 368.4 kj of energy.

Calculate the number of moles of Na in 1.90 g to find the heat released when this quantity reacts.

Atomic mass of Na: 23 g/mol

#mol Na = 1.90 g / 23 g/mol = 0.0826 mol

Do the ratios: [368.4 kj/2mol ] * 0.0826 mol = 15.21 kj.

Then the answer is that 15.21 kj of heat is released (evolved)
6 0
3 years ago
A 4.00L flask containing Ne at 25 C and 6.00 atm is joined by a valve to an 8.00 L flask Ar at 25 C and 2.00 atm.
timama [110]

Answer:

P=3.33atm

Explanation:

Hello!

In this case, since know the volume, temperature and pressure of the initial containers, we can compute the moles of each gas prior to the opening of the valve as shown below:

n_{Ne}=\frac{6atm*4L}{0.08206\frac{atm*L}{mol*K}*298.15K} =0.981molNe\\\\n_{Ar}=\frac{2atm*8L}{0.08206\frac{atm*L}{mol*K}*298.15K} =0.654molAr

Next, we add them up to obtain the total moles:

n_T=0.981mol+0.654mol=1.635mol

Now, the total volume:

V_T=4.00L+8.00L=12.00L

Finally, the total pressure is computed by using the ideal gas equation:

P=\frac{1.635mol*0.08206\frac{atm*L}{mol*K}*298.15K}{12.00L}\\\\P=3.33atm

Best regards!

6 0
3 years ago
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