All categories

3 years ago

How many grams of chlorine gas must react to give 3.52g of BiCl3 according to the equation in exercise 23?

Chemistry

2 answers:

marshall27 [118]3 years ago

6 0

3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl

Brilliant_brown [7]3 years ago

6 0

The correct answer is 1.19 g of chlorine.

The following reaction is:

2Bi (s) + 3Cl₂ (g) ⇒ 2BiCl₃ (s)

In the reaction, it can be witnessed that 3 mol Cl₂ is equal to 2 mol BiCl₃

The molecular weight of BiCl₃ = 315.33

Thus,

3.52 g BiCl₃ = 3.52 g BiCl₃ × 1.00 mol BiCl₃ / 315.33 g BiCl₃

= 0.0112 mol BiCl₃

The mole ratio of Cl₂ and BiCl₃ is,

3 mol Cl₂ / 2 mol BiCl₃

Therefore, the amount of chlorine needed to form 0.0112 mol BiCl₃ is,

0.0112 mol BiCl₃ × 3 mol Cl₂ / 2 mol BiCl₃ = 0.0168 mol Cl₂

Now, the molecular weight of Cl₂ = 70.90

Thus,

0.0168 mol Cl₂ = 0.0168 mol Cl₂ × 70.90 g Cl₂ / 1.00 mol Cl₂

= 1.19 gm Cl₂

Hence, in the mentioned reaction, there is a need of 1.19 g of chlorine to react to produce 3.52 g of BiCl₃.

You might be interested in

9. The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aque

Nezavi [6.7K]

This question is incomplete, the complete question is;

The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aqueous film-forming foam is as follows; 27 41 22 27 23 35 30 33 24 27 28 22 24

( see " use of AFFF in sprinkler systems," Fire technology, 1975: 5)

The system has been designed so that the true average activation time is supposed to be at most 25 seconds.

Does the data indicate the design specifications have not been met?

Test the relevant hypothesis at significance level 0.05 using the P-value approach

Answer:

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

Explanation:

Given the data in the question;

lets consider Null and Alternative hypothesis;

Null hypothesis H₀ : There is sufficient evidence that the true average activation time is at most 25 seconds

Alternative hypothesis H₁ : There is no sufficient evidence that the true average activation time is at most 25 seconds

i.e

Null hypothesis H₀ : μ ≤ 25

Alternative hypothesis H₁ : μ > 25

level of significance σ = 0.05

first we determine the sample mean;

$x^{bar}$ = $\frac{1}{n}$ ∑ $x_{i}$

where n is sample size and ∑ $x_{i}$ is summation of all the sample;

= $\frac{1}{13}$ ( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )

= $\frac{1}{13}$ ( 363

sample mean $x^{bar}$ = 27.9231

next we find the standard deviation

s = √( $\frac{1}{n-1}$ ∑( $x_{i}-x^{bar}$ )²

x ( $x_{i}-x^{bar}$ ) ( $x_{i}-x^{bar}$ )²

27 -0.9231 0.8521

41 13.0769 171.0053

22 -5.9231 35.0831

27 -0.9231 0.8521

23 -4.9231 24.2369

35 7.0769 50.0825

30 2.0769 4.3135

33 5.0769 25.7749

24 -3.9231 15.3907

27 -0.9231 0.8521

28 0.0769 0.0059

22 -5.9231 35.0831

24 -3.9231 15.3907

sum 378.9229

so ∑( $x_{i}-x^{bar}$ )² = 378.9229

∴

s = √( $\frac{1}{13-1}$ ×378.9229 )

s = √31.5769

standard deviation s = 5.6193

now, the Test statistics

t = ( $x^{bar}$ - μ ) / $\frac{s}{\sqrt{n} }$

we substitute

t = ( 27.9231 - 25 ) / $\frac{5.6193}{\sqrt{13} }$

t = 2.9231 / 1.5585

t = 1.88

now degree of freedom df = n - 1 = 13 - 1 = 12

next we calculate p-value

p-value = 0.042299 ( using Execl's ( = TDIST(1.88,12,1)))

Here x=1.88, df=12, one tail

now we compare the p-value with the level of significance

since p-value (0.042299) is lesser than the level of significance ( 0.05)

We Reject Null Hypothesis

Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds

3 0

3 years ago

Draw the major organic product(s) of the reaction of p-bromoaniline with HNO2, HCl.

NARA [144]

Answer:

The major product from this reaction is 4-bromobenzene diazonium chloride.

Explanation:

The reaction of p-bromoaniline with HNO2 and HCl produces diazonium salt. When primary aromatic amines are nitrosated with nitrous acid in the presence of a strong acid such as HCl, diazonium salts are frequently formed. The diazonium salts are a crucial step in the production of halides and azo compounds. The necessary reaction mechanism is depicted in the illustration in the diagram below.