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2 years ago

Which are exmaples of chemical change?

Chemistry

1 answer:

kifflom [539]2 years ago

7 0

Answer:

cold butter melted when heated, lighting a match, baking cookies,

Explanation:

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andrezito [222]

Answer:

yes the one that is circled is correct

Explanation:

4 0

2 years ago

Read 2 more answers

Octane has a density of 0.692 g/ml at 20∘c. how many grams of o2 are required to burn 17.0 gal of c8h18

german

156251.099 grams of O₂ are required to burn 17.0 gal of C₈H₁₈

<h3>Further explanation</h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g / cm³ or kg / m³

Density formula:

$\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}$

ρ = density

m = mass

v = volume

1 gal equal to = 3785.41 ml

then 17.0 gal = 17 x 3785.41 = 64351.97 ml Octane

grams Octane = ρ x ml

grams Octane = 0.692 g.ml x 64351.97

grams Octane = 44531.563

molar mass Octane (C₈H₁₈) = 114

mole Octane = grams : molar mass

mole Octane = 44531.563 : 114

mole Octane = 390.627

From the reaction

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9H₂O

mole C₈H₁₈ : mole O₂ = 1 : 25/2

$mole\:O_2\:=\:\frac{25}{2} \times\:390.627$

mole O₂ = 4882.846

grams O₂ = mole x molar mass

grams O₂ = 4882.846 x 32

grams O₂ = 156251.099

<h3>Learn more </h3>

moles of water

brainly.com/question/1405182

grams of oxygen required

brainly.com/question/8175791

mass of copper required

brainly.com/question/1680090

Keywords: Octane, mole, mass, gal, density

5 0

2 years ago

Which of the following could be considered a scientific statement? (2 points)

Andrej [43]

Answer:

The answer of the scientific STATEMENT is" Appears That Ants Live In Colonies."

7 0

2 years ago

The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does t

Mama L [17]

Answer:

6.1%

Explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law at constant pressure,

$\frac{V_1}{T_1} =\frac{V_2}{T_2}$

$T_2=39\°\ C=39+273=312\ K$

$T_1=59°\ C=273+59=332\ K$

$V_2=V_1\times \frac{T_2}{T_1} \\=V_1\times\frac{312}{332}\\=0.939V_1$

Percentage change in volume

$\%\ change\ in\ volume=\frac{V_1-0.939V_1}{V_1} \times 100\\=6.1\%$

Change in volume of the balloon is 6.1%

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

2 years ago