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2 years ago

Which of the following chemical equations are correctly balanced? Balance the equations that are not balanced.

Chemistry

2 answers:

ira [324]2 years ago

3 0

Answer:

Below:

Explanation:

a)2 Na + F2 = 2 NaF

It's already balanced.

Type is synthesis.

b)H2 + I2 = 2 HI

It's balanced now.

Type's synthesis.

ollegr [7]2 years ago

3 0

Option A

On products side

Na=2
F=2

On reactants side

Na=2
F=2

Hence balanced

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A latex balloon at sea level and room temperature (25°C) has a volume of 2.42 L. When the balloon is released by the child holdi

yarga [219]

Answer:

The answer to your question is: V2 = 1.94 l

Explanation:

Data

V1 = 2.42 l

T1 = 25°C

P1 = 1 atm

V2 = ?

T2 = 25 -11 = 14°C

P2 = 1(0.7) = 0.7 atm

Formula

P1V1/T1 = P2V2/T2

Clear V2 from the equation

V2 = P1V1T2/ P2T1

V2 = (1)((2.42)(14) / (0.7)(25)

V2 = 33.88 / 17.5

V2 = 1.94 l

6 0

3 years ago

Can reactions be more than one of the five types

ehidna [41]

Yes you can it will not just have 5 :)

3 0

3 years ago

Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if

PIT_PIT [208]

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

4 0

3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

6 0

3 years ago