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3 years ago

Calculate the percent composition of C6H12O6

Chemistry

1 answer:

kifflom [539]3 years ago

3 0

Answer:

Did you just let your cat walk on your keyboard? XD

Could you explain this a bit better please?

I'll answer the question in the replies of this comment if you can give me a bit more information.

Explanation:

May I have brainliest please? :)

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What are four everyday items that rely on the behavior of gases to operate properly?

Artist 52 [7]

A stove needs gas to burn (I only have one off the top of my head, sorry :/)

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4 years ago

Which of the following describes the arrangement of valence electrons in a bond between H and F?

Sloan [31]

C; The Valence electrons spend more time around the atom of F

6 0

3 years ago

∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821

Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

6 0

3 years ago

Calculate the ph of a 0.005 m solution of potassium oxide k2o

Alecsey [184]

First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12

3 0

3 years ago

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

4 years ago