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3 years ago

Potassium chloride will dissolve in water but not in oil. Why?

Chemistry

1 answer:

german3 years ago

8 0

Answer:

oil has a lighter liquid weight and is able to keep other chemicals or videmins from dissolving

Explanation:

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A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti

lilavasa [31]

c = 1.14 mol/L; b = 1.03 mol/kg

Molar concentration

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

c = 1.14 mol/1 L = 1.14 mol/L

Molal concentration

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

b = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0

3 years ago

Read 2 more answers

How many total atoms are in 0.290 g of P2O5?

Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert grams of P₂O₅ to moles of P₂O₅.

$\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\$

Step 2. Convert moles of P₂O₅ to molecules of P₂O₅.

$\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\$

$= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\$

Step 3. Convert molecules of P₂O₅ to atoms.

There are seven atoms in 1 mol P₂O₅.

∴ $\text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\$

8 0

4 years ago

How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

• using the molar mass of Ca(OH)2 (74g/mol).

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

• using the molar mass of HNO3 (63.02g/mol).

$4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3$

So, 7.32g of HNO3 are required.

4 0

2 years ago

What is the reduction half-reaction for the following unbalanced redox equation?

bazaltina [42]

First find the oxidation states of the various atoms:
in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 
Note that N gained electrons, ie, was reduced; Cr was oxidized 
Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out 
B.NH4+ → N2 

Which of the following is an oxidation half-reaction? 
A.Sn 2+ →Sn 4+ + 2e- 
Sn lost electrons so it got oxidized

5 0

3 years ago

Read 2 more answers

Anyone know this Please help me

Olin [163]

Answer:

sorry I don't know

Explanation:

have a great time with your family and friends

6 0

3 years ago