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3 years ago

An unknown compound was decomposed into 63.2 g carbon, 5.26 g hydrogen, and 41.6 g oxygen. what is its empirical formula?

1 answer:

4 0

Step 1:
Divide mass of each element with its M.mass in order to find out moles.

C = 63.2 g / 12 g/mol = Moles = 5.26 moles

H = 5.26 g / 1.008 g/mol =  Moles = 5.21 moles

C = 41.6 g / 16 g/mol =  Moles = 2.6 moles

Step 2:
Select moles of the element with least value and divide all moles of element by it,
C H O
5.26/2.6 : 5.21/2.6 : 2.6/2.6

2.02    : 2.00    :    1

Result:
Empirical Formula =    C₂H₂O

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Select the correct balanced oxidation-reduction reaction. Group of answer choices 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+

bezimeni [28]

Answer:

The correct balanced oxidation- reduction reaction is:

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Explanation:

In this reaction, iron (Fe2+) is the reducing agent while Cr2O7^2- is the oxidizing agent.

The ion transfer is represented as shown below:

6 Fe 2+ - 6e- -----------> 6 Fe 3+ (oxidation)

2 Cr^6 + 6e^- ----------> 2 Cr^3 (reduction)

From the unbalanced reaction

Fe2+ + Cr2O72- + H+ ----------> Fe3+ + Cr3+ + H2O we will follow these steps to balance the reaction.

Step 1: break the equation into two half reactions stating which is oxidized and reduced.

Step 2: Balance the atoms on each sides; the hydrogen, oxygen

Step 3: Balance the gain also

Step 4: Give the electron gained on one side to be equal to the electron lost on the other side.

Step 5: Add the two half reactions and simplify the equation.

Doing this, we obtain

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

6 0

2 years ago

A balloon inflated in a room at 27 degrees celsius has a volume of 8.00 L.the balloon is then heated to a temperature of 78 degr

dangina [55]

This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K

V₂ = 9.38 L

5 0

2 years ago

A 125 g compound consists of 37% oxygen? What is the mass of oxygen in this compound?

zepelin [54]

19.44%

I think that's the anwer

7 0

3 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

2 years ago

Five ways you can recycle

Natasha_Volkova [10]

Answer:)

Top 5 Ways to Reuse and Recycle at Home
Repurpose Glass, Plastic and Cardboard Containers. ...
Designate a Kitchen Drawer for Plastic Bags. ...
Reuse your Home Delivered Newspaper. ...
Supply Artists with Creative Materials. ...
Convert Old Sheets, Towels and Clothing into Wash Rags.

Hope it helps you my dear friend:)
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6 0

2 years ago

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