Which statement descibes the type of change and the chemical properties of the product and reactants

Why is bromine more electronegative than iodine?

gavmur [86]

Answer

Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)

8 0

3 years ago

Plzzz help

oee [108]

Answer:

glass

Explanation:

7 0

2 years ago

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Why are predictions of natural hazards important?

vovikov84 [41]

Answer:

Good predictions and warnings save lives.

Explanation:

Predictions and warnings can also reduce damage and economic losses.

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3 years ago

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Consider the reaction of C3H8 with O2 to form CO2 and H2O. If 5.11 g O2 is reacted with excess C3H8 and 3.35 g of CO2 is ultimat

JulsSmile [24]

Answer : The percent yield of the reaction is, 79.8 %

Explanation : Given,

Mass of $O_2$ = 5.11 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $O_2$ .

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{5.11g}{32g/mole}=0.159mole$

Now we have to calculate the moles of $CO_2$ .

The balanced chemical reaction will be,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction, we conclude that

As, 5 moles of $O_2$ react to give 3 moles of $CO_2$

So, 0.159 moles of $O_2$ react to give $\frac{3}{5}\times 0.159=0.0954$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.0954mole)\times (44g/mole)=4.1976g$

The theoretical yield of $CO_2$ = 4.1976 g

The actual yield of $CO_2$ = 3.35 g

Now we have to calculate the percent yield of $CO_2$

$\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{3.35g}{4.1976g}\times 100=79.8\%$

Therefore, the percent yield of the reaction is, 79.8 %

8 0

3 years ago

Be sure to answer all parts. Consider the following equilibrium process at 686°C: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) The equilibriu

Scrat [10]

Answer:

a)The equilibrium constant at 686°C is 0.527.

b) Concentrations of all the gases be when equilibrium is reestablished:

$[CO_2]=0.2834 M$

$[H_2]=0.02440 M$

$[CO]=0.06660 M$

$[H_2O]=0.05460 M$

Explanation:

a) $CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

Equilibrium concentrations:

$[CO]=0.0500 M$

$[H_2]=0.0410 M$

$[CO_2]=0.0880 M$

$[H_2O]=0.0380 M$

The value of an equilibrium constant will be given as:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

$K_c=\frac{0.0500 M\times 0.0380 M}{0.0880 M\times 0.0410 M}=0.527$

The equilibrium constant at 686°C is 0.527.

b)

$CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

Initially

0.30 M 0.0410 M 0.0500 M 0.0380 M

At reestablishment of an equilibrium;

(0.30-x) M (0.0410-x) M (0.0500+x) M (0.0380+x) M

The value of an equilibrium constant will be given as:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

$0.527=\frac{(0.0500+x)M\times (0.0380+x)M}{(0.30-x)M\times(0.0410-x)M}$

Solving for x;

x = 0.0166

Concentrations of all the gases be when equilibrium is reestablished:

$[CO_2]=(0.30-x)=(0.30-0.0166) M=0.2834 M$

$[H_2]=(0.0410-x)=(0.0410-0.0166) M=0.02440 M$

$[CO]=(0.0500+x)=(0.0500+0.0166) M=0.06660 M$

$[H_2O]=(0.0380+x)=(0.0380+0.0166) M=0.05460 M$

7 0

3 years ago