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3 years ago

Complete and balance the equations for the following double displacement reactions:

Chemistry

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

A. AgNO₃ + NaCl —> AgCl + NaNO₃

B. Mg(NO₃)₂ + 2KOH —> 2KNO₃ + Mg(OH)₂

C. 3LiOH + Fe(NO₃)₃ —> Fe(OH)₃ + 3LiNO₃

Explanation:

The equation given in the question above can be balanced as follow.

A. AgNO₃ + NaCl —> AgCl + NaNO₃

The above equation is already balanced.

B. Mg(NO₃)₂ + KOH —> KNO₃ + Mg(OH)₂

There are 2 atoms of H on the right side and 1 atom on the left side. It can be balance by writing 2 before KOH as shown below:

Mg(NO₃)₂ + 2KOH —> KNO₃ + Mg(OH)₂

There are 2 atoms of K on the left side and 1 atom on the right side. It can be balance by writing 2 before KNO₃ as shown below:

Mg(NO₃)₂ + 2KOH —> 2KNO₃ + Mg(OH)₂

Thus, the equation is balanced.

C. LiOH + Fe(NO₃)₃ —> Fe(OH)₃ + LiNO₃

There are 3 atoms of H on the right side and 1 atom on the left side. It can be balance by writing 3 before LiOH as shown below:

3LiOH + Fe(NO₃)₃ —> Fe(OH)₃ + LiNO₃

There are 3 atoms of Li on the left side and 1 atom on the right side. It can be balance by writing 3 before LiNO₃ as shown below:

3LiOH + Fe(NO₃)₃ —> Fe(OH)₃ + 3LiNO₃

Thus, the equation is balanced.

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snow_tiger [21]

The answer my friend is dense and here is why if you compact dense food it becomes even more flat and there is more room for food.

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3 years ago

Read 2 more answers

At STP how many moles or helium would occupy a volume of 12 liters?

butalik [34]

1 mole ------------- 22.4 L ( at STP )
?? mole ---------- 12 L

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hope this helps!

5 0

4 years ago

a rock is examined to determine its age, and the ratio of uranium-235 to lead-207 is found to be 125,000:875,000. the half-life

slamgirl [31]

The old age of rock when examined is 21.4 million years if the ratio of uranium-235 to lead-207 is found to be 125,000:875,000. the half-life of uranium-235 is 700 million years.

<h3>Decaying Equation </h3>

P = [P + D] (1/2) ^(t/t(1/2))

where,

P represent the present parent amount

D is the present daughter amount

t(1/2) is the half life time period

t is the actual age

We know that,

t = (log[(P+D) /P] / log2) × t(1/2) ----------(1)

Given,

P = 97.6

D = 2.1

Ratio of P and D can be calculated as

P/D = 97.6/2.1

= 46.476

By substituting all the values in eq(1), we get

t = [(log 46.476 +1)/log2] × t(1/2)

Given,

The half life of U —Pb decay is 700 million years.

So,

t = [(log 46.476 +1)/log2] × 700

t = 21.4 million years.

Thus, we calculated that the the old age of rock when examined is 21.4 million years.

learn more about Half life period:

brainly.com/question/20309144

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DISCLAIMER:

The given question is misprint on portal.

Here is the correct form of question:

A rock is examined to determine its age, and the ratio of uranium-235 to lead-207 is found to be 125,000:875,000. the half-life of uranium-235 is 700 million years.

How old is the rock if it contains Uranium-235/ lead-207 ratio of 97.6 to 2.1?

8 0

2 years ago

Determine the freezing point of a 0.765 m solution of nitrobenzene in naphthalene. (given: naphthalene Kf = 7.45o C Kg/ mole and

amid [387]

Answer: The freezing point of a 0.765 m solution of nitrobenzene in naphthalene is $74.6^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(80.3-T_f)^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte such as nitrobenzene)

$K_f$ = freezing point constant = $7.45^0C/m$

m= molality = 0.765

$(80.3-T_f)^0C=1\times 7.45\times 0.765$

$T_f=74.6^0C$

Thus the freezing point of a 0.765 m solution of nitrobenzene in naphthalene is $74.6^0C$

7 0

3 years ago

The density of acetonitrile (CH3CN) is 0.786 g/mL, and the density of methanol (CH3OH) is 0.791 g/mL. A solution is made by diss

ololo11 [35]

Answer:

The answer to your question is 0.22

Explanation:

Data

Acetonitrile (CH₃CN) density = 0.786 g/ml

Methanol (CH₃OH) density = 0.791 g/ml

Volume of CH₃OH = 22 ml

Volume of CH₃CN = 98.4 ml

Process

1.- Calculate the mass of Acetonitrile and the mass of Methanol

density = mass/ volume

mass = density x volume

Acetonitrile

mass = 0.786 x 98.4

= 77.34 g

Methanol

mass = 0.791 x 22

= 17.40 g

2.- Calculate the moles of the reactants

Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)

= 24 + 14 + 3

= 41 g

Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)

= 12 + 4 + 16

= 32 g

Moles of Acetonitrile

41 g ----------------- 1 mol

77.34g ------------ x

x = (77.34 x 1) / 41

x = 1.89 moles

Moles of Methanol

32 g -------------- 1 mol

17.40 g --------- x

x = (17.40 x 1)/32

x = 0.54 moles

3.- Calculate the mole fraction of Methanol

Total number of moles = 1.89 + 0.54

= 2.43

Mole fraction = moles of Methanol / total number of moles

Mole fraction = 0.54/ 2.43

Mole fraction = 0.22

6 0

4 years ago