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3 years ago

The Kelvin temperature of sample of 650 cm sample of ammonia gas is doubled what is the new volume of the gas

Chemistry

1 answer:

storchak [24]3 years ago

4 0

Answer:

1300cm3.

Explanation:

Step 1:

Data obtained from the question.

Initial temperature (T1) = K

Initial volume (V1) = 650 cm3

Final temperature (T2) = double the original = 2K

Final volume (V2) =..?

Step 2:

Determination of the new volume of the gas.

The new volume of the gas can be obtained by using Charles' law as follow:

V1/T1 = V2/T2

650/K = V2/2K

Cross multiply

K x V2 = 650 x 2K

Divide both side by K

V2 = 650 x 2K /K

V2 = 650 x 2

V2 = 1300cm3.

Therefore, the new volume of the gas is 1300cm3

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1.20×10−8s to nanoseconds

AveGali [126]

Answer:

There are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

Explanation:

We need to convert $1.2\times 10^{-8}\ s$ to nanoseconds.

We know that,

$1\ s=10^9\ ns$

Now using unitary method to solve it such that,

$1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns$

So, there are 12 nanoseconds in $1.2\times 10^{-8}\ s$ .

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3 years ago

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

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2 years ago

1.What are pollen grain? 2. what is an embryo?<br>

swat32

Answer: 2= baby being developed= embryo 1 i dont know sorry

Explanation:

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3 years ago

Read 2 more answers

Balance the equation <br> Ca(s) + H3PO4(aq) ---->Ca3(PO4)2(s) + H2(g)

vovikov84 [41]

Answer: CO2x+O4H I don’t know if this is the right answer

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2 years ago

Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a

AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

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3 years ago