What types of material might have survived 85 years of weather? answer?

What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua

notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $Na_2S$

Given mass = 15.5 g

Molar mass of $Na_2S$ = 78.0452 g/mol

Moles of $Na_2S$ = 15.5 g / 78.0452 g/mol = 0.1986 moles

Given: For $CuSO_4$

Given mass = 12.1 g

Molar mass of $CuSO_4$ = 159.609 g/mol

Moles of $CuSO_4$ = 12.1 g / 159.609 g/mol = 0.0758 moles

According to the given reaction:

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

1 mole of $Na_2S$ react with 1 mole of $CuSO_4$

0.1986 mole of $Na_2S$ react with 0.1986 mole of $CuSO_4$

Available moles of $CuSO_4$ = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, $CuSO_4$ is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuSO_4$ gives 1 mole of $CuS$

0.0758 mole of $CuSO_4$ gives 0.0758 mole of $CuS$

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

Option B is correct.

6 0

3 years ago

How did planetesimals form?

Nady [450]

Condensation, the process of forming solid particles from the solar nebula, is an early process in the formation of a planetesimal, an early solar system body of small to medium size that combines with otherplanetesimals to form protoplanets

3 0

3 years ago

Read 2 more answers

Calculate the amount of heat needed to boil 64.7 g of benzene (C6H6), beginning from a temperature of 41.9 C . Round your answer

JulsSmile [24]

Answer: The amount of heat needed is = 4.3kJ

Explanation:

Amount of heat H = M × C × ΔT

M= mass of benzene = 64.7g

C= specific heat capacity = 1.74J/gK

ΔT = T2-T1

Where T1 is initai temperature = 41.9C

T2 is the final temperature( boiling point of benzene) = 80.1C

H= 64.7×1.74×80.7

H= 4300J

H=4.3kJ

Therefore, the amount of heat needed is 4.3kJ

8 0

3 years ago

What volume of methanol should be used to make 1.25 L of a 21.0% by volume solution? ___ mL

o-na [289]

Answer:

V CH3OH = 262.5 mL

Explanation:

%v/v = (v compound / v sln)×100

∴ %v/v = 21.0 %

∴ v sln = 1.25 L

⇒ v compound = ?

⇒ 21.0 % = ( v compound / 1.25)×100

⇒ 0.21 = v compound / 1.25 L

⇒ (0.21)*(1.25 L) = v compound

⇒ v compound = 0.2625 L = 262.5 mL

5 0

3 years ago

A student did not read the directions to the experiment properly and mixed up where to place the NaOH solution and the vinegar.

Keith_Richards [23]

Answer:

The answer is "Only at the end, a transformation of rose to color is made ".

Explanation:

In this student puts its vinegar in the titrator, and NaOH throughout the beaker, which implies phenolphthalein has also been poured into water.

Phenolphthalein does have a pH range of 8.3-10 (approx). It's indeed pink in the basic medium therefore, the formulation becomes pink throughout the color by adding phenolphthalein to NaOH.

It is beginning of vinegar was its beginning of neutralization of NaOH from the titrator through full neutralization, a single piece of vinegar is added to a solution as well as the rose solution is lost throughout the beaker.

7 0

3 years ago