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2 years ago

Calculate the Mass of water needed to make 20% solution from 50g of glucose ?

Chemistry

1 answer:

MariettaO [177]2 years ago

3 0

You would need at least 250 grams of water
Because 50g of glucose divided by 250g of water equals 20 percent

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Using the activity series provided. Which reactants will form products? Na > Mg > Al > Mn > Zn > Cr > Fe >

Elodia [21]

Answer:

Fe + Cu(NO3)2 →

Explanation:

3 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

2SO2 + O2 ⇄ 2SO3

Initial 3.4 atm 1.3 atm 0 atm

Change -2x - x + 2x

Equilibrium 3.4 - 2x 1.3 -x 0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

2SO2 + O2 ⇄ 2SO3

Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm

Equilibrium 2.88 atm 1.04 atm 0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

$Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135$

8 0

3 years ago

Describe the theory of ionization in regard to acids and bases

dmitriy555 [2]

<span>Acids are substances which produce hydrogen ions in solution.Bases are substances which produce hydroxide ions in solution.</span>

7 0

3 years ago

Read 2 more answers

During which two processes does a substance release energy?

enyata [817]

Answer:

Freezing and condensation

Explanation:

Because they are exothermic, they release energy

6 0

2 years ago