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2 years ago

Which coefficient correctly balance the equation CaO + H2O -> Ca(OH)2

Chemistry

1 answer:

deff fn [24]2 years ago

4 0

For the equation that we are given:

$CaO+H_{2}O$ → $Ca(OH)_{2}$

We see that on the reactants side we have:

Ca: 1 mol

O: 2 mol

H: 2 mol

And then we see that on the products side we have:

Ca: 1 mol

O: 2 mol

H: 2 mol

Therefore since we have equal moles of each of the elements on both sides of the equation, this equation is balanced how it is.

So the correct coefficients for this equation are: B) 1, 1, 1.

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You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak

Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0

2 years ago

A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and

Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ chloride ions.

In one mole of aldrin there are 6 moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$

8 0

2 years ago

In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are

mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

$W = 200*(1-0.704)\\W=59.2\ g$

There are 59.2 grams of water in this solution.

5 0

2 years ago

At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2

s344n2d4d5 [400]

Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0

2 years ago

In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?

Vinvika [58]

Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, $Au$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms (given)

From the formula of compound that is $KAu(CN)_2$ it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, $K$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms

Number of carbon, $C$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Number of nitrogen, $N$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = Number of gold, $Au$ atoms+Number of potassium, $K$ atoms +Number of carbon, $C$ atoms + Number of nitrogen, $N$ atoms

Total number of atoms in $KAu(CN)_2$ = $6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = $39.96\times 10^{20}$ atoms

Hence, the total number of atoms in $KAu(CN)_2$ is $3.996\times 10^{21}$ atoms.

7 0

2 years ago