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DIA [1.3K]
3 years ago
10

A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad

e of oxygen. You are asked to determine the mass of oxygen in the sample. Which of the following expressions demonstrates a mathematical procedure to solve this problem using the proper order of operations?
Chemistry
1 answer:
Umnica [9.8K]3 years ago
8 0

We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

<span>                                       x = (0.65)(5.5 g)</span>

<span>                                       <em> </em><span><em>x = 3.575 g</em></span></span>

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The diagram below shows part of the rock cycle. (6 points)
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Answer:

Igneous Rock

Explanation:

Assuming this is a cycle, the volcanic eruption would lead back to rock B; rocks formed by volcanic eruptions are considered Igneous.

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An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

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Which compound is a hydrocarbon?
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C2H6 OR C6H12O6 THE ANSWER

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3 years ago
Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re
OLEGan [10]

Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔGo and ΔG have the same magnitude, they just have opposite signs.

Explanation:

The fraction of the total heat energy if a system that does useful work is known as Gibb's free energy (G) and the change from the initial to final state is designated by \Delta G. It is observed that the values of \Delta G changes with experimental conditions such as temperature , pressure , concentration etc.

\Delta G^0 is the standard free energy change which is a balance of two natural tendencies of any system.

  1. Minimization of potential energy or enthalpic factor \Delta H^0

Maximization of disorderliness or entropic factor T\Delta S^0

Mathematically; \Delta G = \Delta H^0 - T\Delta S^0

Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

ΔG⁰ and ΔG can have different values, they don't even have to have the same sign

For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

while the false statements include:

a) If the reaction has a large negative ΔG⁰ value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔG⁰ and ΔG have the same magnitude, they just have opposite signs.

7 0
3 years ago
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