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How does an ion differ from an electrically-neutral atom?

Marysya12 [62]

Answer:

B there are a different number of electrons in an ion compared to a neutral atom

3 0

4 years ago

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In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixt

sergiy2304 [10]

Answer : The percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

And the relation between the rate of effusion and volume is :

$R=\frac{V}{t}$

or, from the above we conclude that,

$(\frac{V_1}{V_2})^2=\frac{M_2}{M_1}$ ..........(1)

where,

$V_1$ = volume of helium gas = 29.7 ml

$V_2$ = volume of mixture = 9.28 ml

$M_1$ = molar mass of helium gas = 4 g/mole

$M_2$ = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

$(\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}$

$M_2=40.97g/mole$

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of $CO$ be, 'x' and the mole fraction of $CO_2$ will be, (1 - x).

As we know that,

$\text{Average molar mass of mixture}=\text{Mole fraction of }CO$

$\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)$

Now put all the given values in this expression, we get:

$40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)$

$x=0.1894$

The mole fraction of $CO$ = x = 0.1894

The mole fraction of $CO_2$ = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of $CO$ = $0.1894\times 100=18.94\%$

The percent composition by volume of mixture of $CO_2$ = $0.8106\times 100=81.06\%$

Therefore, the percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

6 0

3 years ago

Suppose six atoms of oxygen are present at the beginning of a chemical reaction. How many oxygen atoms are present when the reac

wariber [46]

The answer should be a) always six. The chemical reaction is the process of molecule break down and form again. So the molecule will change but the atom will not change both type and number.

3 0

3 years ago

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What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?

Marat540 [252]

Given:

% Al = 35.94

% S = 64.06

To determine:

Empirical formula of a compound with the above composition

Explanation:

Atomic wt of Al = 27 g/mol

Atomic wt of S = 32 g/mol

Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g

# moles of Al = 35.94/27 = 1.331

# moles of S = 64.06/32 = 2.002

Divide by the smallest # moles:

Al = 1.331/1.331 = 1

S = 2.002/1,331 = 1.5 ≅ 2

Empirical formula = AlS₂

6 0

3 years ago

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D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match th

Aleks04 [339]

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

6 0

3 years ago