Heart attacks can cause sudden death.<br><br> True <br><br> False

Which of the following compounds will not produce O2 gas when it decomposes? (2 points)

Umnica [9.8K]

LiOH is the right answer.

6 0

3 years ago

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

3 years ago

What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization

zzz [600]

Answer:

$\boxed{\text{889 g}}$

Explanation:

The formula relating the mass m of a sample and the heat q to vaporize it is

q = mL, where L is the latent heat of vaporization.

$\begin{array}{rcl}2000 \times 10^{3} \text{ J} & = & m \times \dfrac{2.25 \times 10^{6} \text{ J}}{\text{1 kg}}\\\\m & = & \dfrac{2000 \times \times 10^{3}\text{ kg}}{2.25 \times 10^{6}}\\ & = & \text{0.889 kg}\\\\ & = & \text{889 g}\\\end{array}\\\text{The mass of water is $\boxed{\textbf{889 g}}$}$