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2 years ago

Which of the following could be an empirical formula? a C6H12O6 b H2O2 c CH2O d C6H6

Chemistry

2 answers:

Crank2 years ago

5 0

Answer:

A C6H12O6 will be the empirical formulae.

AfilCa [17]2 years ago

4 0

I dont know i just want help thats why im answering to get 1 help as i have to help to get help

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Can someone help? I wasn't at school the day we did this and I don't understand.

Anton [14]

1) 2700 kg/l
2) 13.6 kg/l
3) 0.1578 kg
4) 8921.5 kg/m3
5) 1.59 kg/l
6) 1.84 kg/l
7) 0.21965 kg
8) 11331.9 kg/m3
9) 7.9167 kg/l
10) 238.095 cm3

Just divide the masses by volume to find out the density, multiply the volume with density to find out the mass and divide the mass by density to find out the volume.
To turn the result into SI unit (kg/l), divide the g by 1000 and ml by 1000.

7 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Dovator [93]

Answer: 406 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000$ Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = $\frac{193000}{63.5}\times 19000=57748032$ Coloumb

$57748032=39.5\times t$

$t=1461975sec=406hours$ (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

3 0

3 years ago

Classify the outcomes based on whether they are caused by adding or removing a reactant from a chemical reaction. The rate of th

Mama L [17]

Explanation: To study the outcomes, we will apply Le-Chatelier's principle.

Le-Chatelier's principle states that if there is any disturbance in the conditions of the dynamic equilibrium, the position of equilibrium will counteract the change.

1) The rate of the forward reaction increases.

This will happen when we add the reactant to a chemical reaction. According to Le-Chatelier's principle, by increasing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, forward reaction is favored.

2) The rate of the revere reaction increases.

This will happen when we remove the reactant from a chemical reaction. According to Le-Chatelier's principle, by removing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, reverse reaction is favored.

3) The concentration of product increases.

This will happen when we add reactants to a chemical reaction. According to Le-Chatelier's principle, when we increase the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the forward direction which means that the concentration of product will increase.

4) The concentration of products decreases.

This will happen when we remove reactants from a chemical reaction. According to Le-Chatelier's principle, when we decrease the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the reverse direction which means that the concentration of reactants will increase or concentration of products will decrease.

3 0

3 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago

A container is at a pressure of 3 atm and a temperature of 280K. What is the new temperature when the pressure is reduced to 1.5

GenaCL600 [577]

Answer:

140 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 3 atm

Initial temperature of the gas (T₁): 280 K

Final pressure of the gas (P₂): 1.5 atm

Final temperature of the gas (T₂): ?

Step 2: Calculate the final temperature of the gas

We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.

T₁/P₁ = T₂/P₂

T₂ = T₁ × P₂/P₁

T₂ = 280 K × 1.5 atm/3 atm = 140 K

6 0

3 years ago