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sammy [17]
3 years ago
7

What is the ratio of carbon, hydrogen, and oxygen in most carbohydrates?

Chemistry
2 answers:
horrorfan [7]3 years ago
7 0
 In carbohydrates the C:H:O is 1:2:1
san4es73 [151]3 years ago
4 0

The ratio is— 1:2:1 (carbon, hydrogen, oxygen)

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The equation below represents a double replacement/displacement reaction;
Naddik [55]
AB+XY->AY+XB
We know that the answer would be KCl because of the switching that takes place during a double displacement reaction. Just like Zn and MnO4 switched and combined, the remaining elements, K and Cl, will combine.
We know that the answer is simply KCl because both K and Cl have an ion of only +/-1, meaning when they cross, no suffixes are made, since their ions are only 1.
For example, if you combined Mg with Cl, you would get MgCl2, because Mg has an ion of +2.
I hope this helps!
4 0
3 years ago
Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn
MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

  • CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒  ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0
3 years ago
Read 2 more answers
Oxidation no of carbon in C3H7OH​
mestny [16]

Answer:

-2

Explanation:

7 x 1 - 2 x 1 + 1 x 1 + 3C = 0 (no charge)

6 + 3C = 0

C = -2

6 0
4 years ago
Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to<br> neutralize it?
Sonbull [250]
In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.
3 0
3 years ago
A sample of a gas has a volume of 852 mL at 298 K. If the gas is cooled to 200K, what would the new volume be?
Colt1911 [192]

Answer:

571.81 mL

Explanation:

Assuming constant pressure, we can solve this problem by using <em>Charles' law</em>, which states that at constant pressure:

  • V₁T₂=V₂T₁

Where in this case:

  • V₁ = 852 mL
  • T₂ = 200 K
  • V₂ = ?
  • T₁ = 298 K

We <u>input the data</u>:

  • 852 mL * 200 K = V₂ * 298 K

And <u>solve for V₂</u>:

  • V₂ = 571.81 mL

The new volume would be 571.81 mL.

4 0
3 years ago
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