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saveliy_v [14]
3 years ago
14

How many carbon atoms are in 6.5 moles of carbon?

Chemistry
1 answer:
musickatia [10]3 years ago
6 0
Just do 6.5 * (6.02 * 10^23)
Sorry I don’t have a calculator on me
Type that into the calculator to get your answer
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In the context of small molecules with similar molar masses, arrange these intermolecular forces by strength (hydrogen bonding -
Katarina [22]
<h2>Answer:</h2>

Arrangement of inter molecular forces from strongest to weakest.

  • Hydrogen bonding
  • Dipole-dipole interactions
  • London dispersion forces.
<h3>Explanation:</h3>

Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.

Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.

Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.

But they are weaker than hydrogen bonding.

London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.

5 0
3 years ago
What is this equation?
DIA [1.3K]

2C_{7}H_{10} + 19O_{2} → 14CO_{2} + 10H_{2}O

8 0
2 years ago
HELP ME PLEASE (brainliest)
Aloiza [94]

Answer: i would say D or the last one.

Explanation: According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

Hope this helps :) Can u plz mark me branliest

7 0
2 years ago
The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is
Feliz [49]
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
5 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
3 years ago
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