A compound formed with potassium and bromine that can detect blood in a stain by reacting with it to form distinctive crystals

Which of the following conversion factors is correct to convert 0.25 moles of Na to particles of Na?

Stolb23 [73]

Answer:

6.02×10²³ particles / 1 mol

Explanation:

In this question the choices are missed.

For determine the particles of Na, in 0.25 moles we should know that the mole referrs to the Avogadro's number.

1 mol of anything contains 6.02×10²³ entities, that's why we should apply this conversion factor to find the final answer:

0.25 mol . 6.02×10²³ particles / 1 mol = 1.50×10²³ particles

In conclussion: 1.50×10²³ particles of Na are contained in 0.25 moles of Na

5 0

4 years ago

If 5.60 mol of a substance releases 14.9 kJ when it decomposes, what is ΔH for the process in terms of kJ/mol of reactant?

lisov135 [29]

Answer:

The correct approach is "-2.67 kJ/mole". A further solution is described below.

Explanation:

The given values are:

No. of moles,

= 5.60 mol

Substance releases,

ΔH = 14.9 kJ

Now,

The ΔH in terms of kJ/mol will be:

= $\frac{\Delta H}{5.60}$

On putting the given values of ΔH, we get

= $\frac{-14.9}{5.60}$

= $-2.67 \ kJ/mole$

8 0

3 years ago

An unknown compound was found to have a percent composition as follows: 47.0%, 14.5 carbon, and 38.5 oxygen. What is its empiric

shusha [124]

KCO₂

K₂C₂O₄

Explanation:

Given parameters:

Percent composition:

K = 47%

C = 14.5%

O = 38.5%

Molar mass of compound = 166.22g/mol

Unknown:

Empirical formula of compound = ?

Molecular formula of compound = ?

Solution:

The empirical formula of a compound is its simplest formula. Here is how to solve for it:

K C O

Percent

composition 47 14.5 38.5

Molar mass 39 12 16

Number

of moles 47/39 14.5/12 38.5/16

moles 1.205 1.208 2.4

Dividing

by smallest 1.205/1.205 1.208/1.205 2.4/1.205

1 1 2

Empirical formula KCO₂

Molecular formula

This is the actual combination of the atoms:

Molecular formula = ( empirical formula of KCO₂)ₙ

Molar mass of empirical formula = 39 + 12 + 2(16) = 83g/mol

n factor = $\frac{true molecular mass}{molar mas of empirical formula}$

n factor = $\frac{166.22}{83}$ = 2

Molecular formula of compound = ( KCO₂)₂ = K₂C₂O₄

Learn more:

Empirical formula brainly.com/question/2790794

#learnwithBrainly

6 0

4 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

3 years ago

Salts should generally be soluble in nonpolar organic solvents. <br> a. True <br> b. False

Rashid [163]

I think its A but im not completely sure.

3 0

3 years ago

Read 2 more answers