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Ulleksa [173]
4 years ago
10

How do hydrocarbons differ in structure from carbohydrates.

Chemistry
1 answer:
ICE Princess25 [194]4 years ago
4 0
Hydrocarbons are carbon and hydrogen. Methane is CH4, and propane is C3H8. Methene is CH3, and propene is C3H6. Carbohydrates are hydrates of carbon. They <span>have the general formula (CH2O)x. thats how it differs
</span>
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Will a precipitate form when 20.0 ml of 0.10 M Ba(NOxaq) and 50.0 mL of 0.10 M NaCO(aq) are mixed together?
Tom [10]

Answer:

B. Q > K precipitate will form

Explanation:

The reaction is;

Ba(NO3)2(aq) + Na2CO3(aq) ------> BaCO3(s) + 2NaNO3(aq)

Hence the reaction could form a precipitate of BaCO3.

Number of moles of carbonate ions = 50/1000 * 0.10 M =  5 * 10^-3 moles

Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles

Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L

Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M

Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M

Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3

But K = 2.58  ×  10 ^− 9

We can clearly see that Q>K therefore precipitate will form

5 0
3 years ago
URGENT PLEASEEEE, WILL GIVE 100 POINTS
Amanda [17]

Let's balance

  • 2N_2O+3O_2---->4NO_2

Moles of laughing gas

  • 9/44
  • 0.2mol

2 moles need 3 mol O_2

1 mol needs= 1.5mol O_2

moles of O_2.

  • 1.5(0.2)
  • 0.3mol

#2

  • 3 molO_2 produces 4 mol smog
  • 1 mol produces 4/3=1.3mol smog

Moles Of O_2

  • 7.5/32
  • 0.2mol

Moles of smog

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Mass

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3 0
2 years ago
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lord [1]

Answer:

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5 0
3 years ago
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All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxyg
tankabanditka [31]

Answer:

High partial pressure of oxygen in the tissues.

Explanation:

Hemoglobin is a protein found in the red blood cells that is responsible for the transportation of oxygen from the lungs to the rest of the tissues in the body. The partial pressure determines whether oxygen is loaded or unloaded into the hemoglobin.

When there is lack of oxygen in our bodies, we experience hypoxia.

6 0
3 years ago
A tank of gas is found to exert 8.6 atm at 38°C. What would be the required
Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0
3 years ago
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