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stepladder [879]
4 years ago
11

In which laboratory process could a student use 0.10 M NaOH(aq) to determine the concentration of an aqueous solution of HBr?

Chemistry
2 answers:
telo118 [61]4 years ago
8 0
(4) titration, the process of using an aqueous solution of known concentration to determine the concentration of another solution of unknown concentration, is your answer.
RideAnS [48]4 years ago
7 0

Answer:

4) titration

Explanation:

Titration is a standard process used in a laboratory to determine the concentration of an unknown analyte. A titrant of known concentration is gradually added to a known volume of the analyte in the presence of a suitable indicator. The end of the titration is marked by a color change of the analyte.

The given example is that of an acid(HBr) - base(NaOH) titration which can be represented by the following equation:

NaOH + HBr → NaBr + H2O

Thus  1 mole of acid gets neutralized by 1 mole of the base to form 1 mole of the salt (NaBr)

Let M1 and V1 are the molarity and volume of the base (NaOH). Here, the molarity of NaOH is known = M1 =  0.10 M and the volume, V1 corresponds to the end point in the titration.

M2 and V2 are the molarity and volume of HBr. Here, V2 is  known whereas M2 needs to be determined.

Based on the reaction stoichiometry:

moles of NaOH = moles of HBr

M1*V1=M2*V2\\\\Therefore,\\\\M2 = \frac{M1*V1}{V2}

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Nitrogen gas in an expandable container is cooled from 45.0 C to 12.3 C with the pressure held constant at 2.58 ∗ 105 Pa. The to
Tresset [83]

Answer:

The number of moles of the gas is: -27.14 mole

the charge in internal energy of the gas is -1.84 × 10⁴ J.

The work done by the gas is 4.42 × 10⁴ J

The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

Explanation:

The expression for the number of moles of a gas at constant pressure is as follows:

\mathbf{n = \frac{Q}{Cp \Delta T}}

\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}

where ;

C_p is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K

Since heat is liberated from the gas ; then:

n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}

n = -27.14 mole

The number of moles of the gas is: -27.14 mole

b) The expression to be used in order to determine the change internal energy is:

dU = nCv \Delta T

where ;

n= 27.14 mole

Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K

ΔT = (12.3-45)

So;

dU = (27.14)(20.76)(12.3-45)

dU = 563.426(-32.7)

dU = -18424.04328

dU = -1.84 × 10⁴ J

Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.

c)  The workdone by the gas can be calculated as;

W = Q - ΔU

W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )

W = 2.58  × 10⁴ J + 1.84  × 10⁴ J

W = 4.42 × 10⁴ J

The work done by the gas is 4.42 × 10⁴ J

d) The expression to calculated the work done is given as:

W = pdV

since the volume is given as constant ; then dV = 0

so;

W = p(0)

W = 0

Replacing 0 for W in the equation W = Q - Δ U

0 = Q - ΔU

-Q = - ΔU

Q = ΔU

Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

5 0
3 years ago
Read 2 more answers
If 45.0 g of O2 are mixed with 45.0 g of H2 and the mixture is ignited, what mass of water is produced?
lbvjy [14]

Answer:

<h2><em><u>45.0g</u></em></h2>

Explanation:

hope it will help you

mark me a brilliant

7 0
3 years ago
What mass of oxygen will react with 2.64 g of magnesium?<br><br> 2Mg(s) + O2(g) → MgO(s)
AleksandrR [38]

Answer:

(24 \times 2) \: g  \: of \: magnesium \: reacts \: with \: (16 \times 2) \: g \: of \: oxygen \\ 2.64 \: g \: of \: magnesium \: will \: react \: with \: ( \frac{2.64 \times 16 \times 2}{24 \times 2} ) \: g \\  = 1.76 \: g \: of \: oxygen

8 0
3 years ago
Uranium can be isolated from its Ores by dissolving it as UO2(NO3)2, then separating If as solid UO2(C2O4).3H2O. Addition of 0.4
Arada [10]

Answer:

The limiting reactant is NaC₂O₄ and the yield of this reaction is 69.52%.

Explanation:

NaC₂O₄ + UO₂(NO₃)₂ + 3H₂O → UO₂(C₂O₄).3H₂O + 2NaNO₃

m (NaC₂O₄) = 0.4031 g  MW = 134 g/mol ∴ n = 3.0 mmol

m (UO₂(NO₃)₂) = 1.481 g MW = 396.05 g/mol ∴ n = 3.74 mmol

m (UO₂(C₂O₄).3H₂O) = 1.073 g MW = 412.094 g/mol ∴ n = 2.60 mmol

The limiting reactant is NaC₂O₄

The yield can be given by (2.60/3.74).100% = 69.52%

6 0
3 years ago
Read 2 more answers
Given the following unbalanced thermochemical equation, how much heat will be released from the combustion of 45.5 grams of CH4
AnnyKZ [126]

CH4 + 2O2 → CO2 + 2H2O + 890 kJ

MM of CH4 = (12.01 + 4x1.008) g/mol = 16.04 g/mol

Moles of CH4 = 45.5 g CH4 x (1 mol CH4/16.04 g CH4) = 2.837 mol CH4

q = 2.837 mol CH4 x (890 kJ/1 mol CH4) = 2520 kJ

8 0
4 years ago
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