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2 years ago

11

what is the greatest amount of bread rolls that can be made with 6 cups of flour And 3 cups of water? If this were a chemical re

action what would Be a limiting reactant? Which reactant is in excess, and how much is leftover?

1 answer:

forsale [732]2 years ago

7 0

Answer:

if you tell me how much is needed and how much you have then i can answer it, but there is not enough information provided to answer to that question.

Explanation:

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RUDIKE [14]

Answer:

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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:

1)

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 $1.2\times 10^{-4}$ .

2)

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 $1.6\times 10^{-10} M$ .

3.

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

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

18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

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3 years ago

If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure

CaHeK987 [17]

Answer : The volume of gas will be 29.6 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 12 atm

$P_2$ = final pressure of gas = 14 atm

$V_1$ = initial volume of gas = 23 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 200K

$T_2$ = final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{12atm\times 23L}{200K}=\frac{14\times V_2}{300K}$

$V_2=29.6L$

Therefore, the new volume of gas will be 29.6 L

5 0

2 years ago

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