1

Type the correct answer in the box. Express your answer to two significant figures. An industrial vat contains 650 grams of soli

nikklg [1K]

88.98 %

The Balance Chemical Equation is as follow,

2 HCl + Pb(NO₃)₂ → 2 HNO₃ + PbCl₂

According to equation,

331.2 g (1 mole) Pb(NO₃)₂ produces = 278.1 g (1 mole) PbCl₂

So,

870 g of Pb(NO₃)₂ will produce = X g of PbCl₂

Solving for X,

X = (870 g × 278.1 g) ÷ 331.2 g

X = 730.5 g of PbCl₂

Therefore,

Theoretical Yield = 730.5 g

Also as given,

Actual Yield = 650 g

So using following formula for percentage yield,

%age Yield = (Actual Yield / Theoretical Yield) × 100

Putting values,

%age Yield = (650 g / 730.5 g) × 100

%age Yield = 88.98 %

Brianliest please and thank you.

6 0

4 years ago

Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and Tr

ale4655 [162]

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

$=\frac{x^2}{0.02 -x}$

$=8.32 \times 10^{-9}$

Clearing $x$ , we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$ , one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

$= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.

3 0

3 years ago

In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al

tekilochka [14]

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)$

here underlined is $HCH_3CO_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

Here underlined is $NH_3$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

Here underlined is $C_2H_5NH_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)$

As $HCH_3CO_2(aq)$ is donating a proton , it acts as a bronsted acid.

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

As $NH_3$ contains a lone pair of electron on nitrogen , it can easily donate electrons to $BH_3$ and act as lewi base.

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

As $C_2H_5NH_2(aq)$ is accepting a proton , it acts as a bronsted base.

7 0

3 years ago

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

Which of the following is the smallest volume? (2 points) 2500 mL

Virty [35]

So to put them all in the same units we have
2500 mL 
250 mL 
25mL 
2,500,000,000mL 

So the third one is the smallest

6 0

4 years ago

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