Ask question

Ask question

All categories

3 years ago

10

What is the mass of 1.72 moles of sodium nitrate? Use the periodic table and the polyatomic ion resource. A. 85.0 g B. 91.2 g C.

146 g D. 273 g

2 answers:

Mrac [35]3 years ago

7 0

The answer is C. 146g because you add all of the masses of the individual elements and then mulyiply by 1.72 to get your answer.

saveliy_v [14]3 years ago

7 0

The Answer is 146 grams (C.)

You might be interested in

What is the most common acid and base on earth?

nata0808 [166]

Answer:

hydrochloride acid

Explanation:

hope this helps :D

5 0

3 years ago

Describe the sequence of events in the formation of an evaporite

erica [24]

- Water enter from creeks, bringing whatever material it has dissolved in its way

- The materials is being build up throughout time in the lake or sea, causing its salinity to increase

- when this saline solution is evaporated, the ions in the water precipitate, forming an evaporite
<span />

3 0

3 years ago

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

c)

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

d)

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

2 years ago

Question 6 (1 point)

Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0

3 years ago

How molecules of N2 gas can be present in a 2.5 L flask at 50°C and 650 mmHg?

ratelena [41]

Answer:

0.482 ×10²³ molecules

Explanation:

Given data:

Volume of gas = 2.5 L

Temperature of gas = 50°C (50+273 = 323 k)

Pressure of gas = 650 mmHg (650/760 =0.86 atm)

Molecules of N₂= ?

Solution:

PV= nRT

n = PV/RT

n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k

n = 2.15 atm. L /26.52 atm. mol⁻¹.L

n = 0.08 mol

Number of moles of N₂ are 0.08 mol.

Number of molecules:

one mole = 6.022 ×10²³ molecules

0.08×6.022 ×10²³ = 0.482 ×10²³ molecules

5 0

3 years ago