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SCORPION-xisa [38]
3 years ago
12

What is the molar mass of NaCl (table salt)?

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
8 0
Na = 22.99 amu
Cl = 35.45 amu

NaCl = 22.99 + 35.45 => 58.44 g/mol 
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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T
Morgarella [4.7K]

Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

7 0
3 years ago
Ozone (O 3) is made of three atoms of oxygen. It is _
kykrilka [37]
An element I think maybe
3 0
3 years ago
Read 2 more answers
10. Copper(i) bromide reacts with magnesium metal: 2 CuBr + Mg → 2 Cu + MgBrz
velikii [3]

Answer:

72.6 grams

Explanation:

I got this answer through stoichiometry.  For every 1 mole of Mg, 2 moles of CuBr are consumed.  Because of this, multiply the moles of Mg by 2.  Then, convert moles to grams.

5 0
3 years ago
Read 2 more answers
An antacid tablet with an active ingredient of CaCO3 was dissolved in 50.0 mL of 0.300 M HCl. When this solution was titrated wi
yaroslaw [1]

Answer:

Explanation:

mole of HCl remaining after reaction with CaCO₃

= .3 M of NaOH of 32.47 mL

= .3 x .03247 moles

= .009741 moles

Initial HCl taken = .3 x .005 moles = .0015 moles

Moles of HCl reacted with CaCO₃

= .009741 - .0015 = .008241 moles

CaCO₃     +    2HCl   =   CaCl₂  +  CO₂  +  H₂O .

1 mole        2 moles

2 moles of HCl reacts with 1 mole of  CaCO₃

.008241 moles of HCl reacts with .5 x .008241 moles of  CaCO₃

CaCO₃ reacted with HCl =  .5 x .008241 = .00412 moles

the mass (in grams) of calcium carbonate in the tablet

= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )

5 0
2 years ago
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