What is the electron configuration of Zn2+

How many grams are in 3.14 moles of PI₃?

OverLord2011 [107]

Answer:

$\boxed {\boxed {\sf 1290 \ g \ PI_3}}$

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

Phosphorus (P): 30.973762 g/mol
Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

I₃: 126.9045 * 3=380.7135 g/mol

PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

$\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

We want to convert 3.14 moles to grams, so we multiply by that value.

$3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

The units of moles of PI₃ cancel.

$3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }$

$1292.698 \ g\ PI_3$

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

1292.698

The 2 in the ones place tells us to leave the 9.

$1290 \ g \ PI_3$

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0

2 years ago

Which of the following describes the correct order for using the scientific

larisa [96]

A. is the answer
if u have any doubts ask me
please mark as brainliest

8 0

3 years ago

Read 2 more answers

The temperature in Jersey City, NJ in mid-Spring is 60*F. The temperature in Miami, FL is 90*F on the same day. Miami is about 1

timurjin [86]

Answer:

Temperature gradient = 30

90F - 60F = 30F

The temperature gradient is 30F.

If I am right, let me know.

7 0

2 years ago

Read 2 more answers

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

4. The type of thermal energy transfer that heats our homes is called

andrew-mc [135]

Answer:

conduction is the answer

6 0

2 years ago