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2 years ago

An indicator is_____ Select all that apply.

Chemistry

1 answer:

crimeas [40]2 years ago

5 0

Answer: C

Explanation:

Indicators are organic weak acids or bases with complicated structures.

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What is the name of this alkane?

Otrada [13]

It's a cyclohexane ring with an ethyl group at 1 and a methyl group at 3. The Ethyl group is bigger and more important group get's the first position.

1-ethyl-3-methylcyclohexane

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3 years ago

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Consider an ideal gas at 30 ∘C and 1.02 atm pressure. To get some idea how close these molecules are to each other, on the avera

Nastasia [14]

Answer: They are close to each other by 41.03 m^3

Explanation:

From Ideal gas equation, PV = nRT

Where n is negligible

R is gas constant = 8.314 J/mol.k

T = 30 + 273 = 303K

P = 1.02 * 103351.5 = 103351.5 Pascal

Then;

PV = RT

V = P/RT

V = 103351.5/(8.314*303)

V = 41.03m^3

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3 years ago

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Ocean basins will expand as new oceanic crust forms and moves away from a mid-oceanic ridge during

KATRIN_1 [288]

Sea floor spread ing

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3 years ago

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a flask contains 2 gases. the first gas has a pressure of 0.76 atm and the second gas gas a pressure of 4,500 Pa. what is the to

Liula [17]

Answer:

The total pressure is 0,804 atm

Explanation:

We use Dalton's law according to which the sum of the partial pressures is equal to the total pressure of a gas mixture. We convert the pressure in Pascals to atmosphere (it can also be done in reverse):

101300Pa ----1 atm

4500Pa----x= (4500Pa x 1atm)/101300Pa= 0,044 atm

P total= p1 + p2= 0,76 atm + 0,044 atm=0,804 atm

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2 years ago

A 29.7 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to

Softa [21]

<u>Answer:</u> The mass of iron in the ore is 10.9 g

<u>Explanation:</u>

We are given:

Mass of iron (III) oxide = 15.6 g

We know that:

Molar mass of Iron (III) oxide = 159.69 g/mol

Molar mass of iron atom = 55.85 g/mol

As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.

To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:

In 159.69 g of iron (III) oxide, mass of iron present is $(2\times 55.85)=111.7g$

So, in 15.6 g of iron (III) oxide, mass of iron present will be = $\frac{111.7g}{159.69g}\times 15.6g=10.9g$

Hence, the mass of iron in the ore is 10.9 g

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3 years ago