Which activity would contribute the most to water pollution?

PLEASE HELP: You have been given 2.42 moles of beryllium sulfide (BeS), determine the mass in grams of beryllium sulfide that yo

Alecsey [184]

The mass in grams of the given beryllium sulfide is 99.2 g. The correct option is the second option - 99.2 g BeS

<h3>Calculating mass of a compound </h3>

From the question, we are to determine the mass of the given beryllium sulfide.

From the given information,

Number of moles of beryllium sulfide (BeS) given = 2.42 moles

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of BeS = 41 g/mol

Then,

Mass of BeS = 2.42 × 41

Mass of BeS = 99.22

Mass of BeS ≅ 99.2 g

Hence, the mass in grams of the given beryllium sulfide is 99.2 g. The correct option is the second option - 99.2 g BeS

Learn more on Calculating mass of a compound here: brainly.com/question/18142599

3 0

3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago

What are three limitations of current cloaking technology.

kirill [66]

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

4 0

3 years ago

If NaCl has a mass of 3.2g, what is the volume of chlorine gas at STP?

Shalnov [3]

Hey there!

Molar mass NaCl = 58.44 g/mol

58.44 g ----------------- 22.4 ( at STP )

3.2 g -------------------- Volume ??

Volume = ( 3.2 x 22.4 ) / 58.44

Volume = 71.68 / 58.44

Volume = 1.226 L

Hope this helps!

5 0

3 years ago

Read 2 more answers

Explain how to separate sugar from supersaturated sugar solution<br> guys pls help

brilliants [131]

A “supersaturated” solution contains more dissolved material. supersaturated solutions lies in the temperature of the water. more sugar will dissolve in hot water than in cold. Meaning that by separating the 2, only the supersaturated sugar would dissolve leaving the regular sugar untouched.

5 0

3 years ago