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3 years ago

List 4 factors that suggest a chemical reaction has taken place

1 answer:

3 0

The change in smell, The change in taste, If there is rust, i cant think of any other on the spot but if something were to catch fire then there is a chemical change.

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I need the answer so much

Mariulka [41]

I don't know sorry just google it

4 0

3 years ago

CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o

Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

$CO_{(g)}$ + $2H_2O_{(g)}$ ⇄ $CH_3OH_{(g)}$

The ICE table can be represented as:

$CO_{(g)}$ + $2H_2_{(g)}$ ⇄ $CH_3OH_{(g)}$

Initial 0.27 0.49 0.0

Change -x -2x x

Equilibrium 0.27 - x 0.49 -2x x

We can now say that the concentration of $CH_3OH_{(g)}$ at equilibrium is x;

Let's not forget that at equilibrium $CH_3OH_{(g)}$ = 0.11 M

So:

x = [ $CH_3OH_{(g)}$ ] = 0.11 M

[ $CO_{(g)}$ ] = 0.27 - x

[ $CO_{(g)}$ ] = 0.27 - 0.11

[ $CO_{(g)}$ ] = 0.16 M

[ $2H_2_{(g)}$ ] = (0.49 - 2x)

[ $2H_2_{(g)}$ ] = (0.49 - 2(0.11))

[ $2H_2_{(g)}$ ] = 0.49 - 0.22

[ $2H_2_{(g)}$ ] = 0.27 M

$K_C = \frac{[CH_3OH]}{[CO][H_2]^2}$

$K_C = \frac{(0.11)}{(0.16)[(0.27)^2}$

$K_C = 9.4307$

$K_C$ = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0

3 years ago

Which energy source has been developed and improved by chemists?

Nostrana [21]

Nuclear hope this helps

8 0

3 years ago

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Sarah performs an experiment to see whether bees prefer red flowers or yellow flowers. Look at this table of results from the ex

denis-greek [22]

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3 years ago

A chemist prepares a solution by dissolving 0.3 mole of CaCl2 in enough water to make 1000 mL of solution. What is the molar con

Rasek [7]

$CaCl_{2}$ dissolves in water to give $Ca^{2+}$ and $Cl^{-}$ ions according to the following reaction:

$CaCl_{2}$ --------> $Ca^{2+}$ + 2 $Cl^{-}$

So, according to the above reaction, 1 mole of $CaCl_{2}$ produce 2 moles of $Cl^{-}$ ion,

So, 0.3 mole will give = 0.3 x 2 = 0.6 moles of $Cl^{-}$ ion

So, Molar concentration = $\frac{0.6moles}{1L} = 0.6$

Note: 1L = 1000mL

7 0

3 years ago