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Zolol [24]
2 years ago
15

The elements X and Y combine in different ratios to form four different types of compounds: XY, XY2, XY3, and XY4. Consider that

there is enough of each compound to contain 2 g of X. In XY the mass of X is 2 g and the mass of Y is 4 g.
Arrange the following ratios in order of their increasing value. Rank from highest to lowest ratio. To rank items as equivalent, overlap them.

1. The ratio of the mass ratio of Y to X in XY3 to the mass ratio of Y to X in XY.
2. The ratio of the mass ratio of Y to X in XY2 to the mass ratio of Y to X in XY.
3. The ratio of the mass ratio of Y to X in XY4 to the mass ratio of Y to X in XY.

Chemistry
1 answer:
Law Incorporation [45]2 years ago
3 0

Answer:

Ratios in order of increasing value ; The ratio of the mass ratio of Y to X in XY2 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY3 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY4 to the mass ratio of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

The detailed and step by step calculation is shown in the attachment.

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Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is
Citrus2011 [14]
Depression in freezing point (ΔT_{f}) = K_{f}×m×i,
where, K_{f} = cryoscopic constant = 1.86^{0} C/m,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For NaNO_{3})

Thus, (ΔT_{f}) = 1.86 X 0.0085 X 2 = 0.03162^{0}C

Now, (ΔT_{f}) = T^{0} - T
Here, T = freezing point of solution
T^{0} = freezing point of solvent = 0^{0}C
Thus, T = T^{0} - (ΔT_{f}) = -0.03162^{0}C
8 0
2 years ago
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A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of sulfur, and 64.00 g of oxygen. How many grams of sulfur and grams of o
Gemiola [76]
From the chemical formula of sulfuric acid, we can see the molar ratio:

H : S : O 
2 : 1 : 4

Now, we convert the mass of hydrogen given into the moles of hydrogen. This is done using

Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27

Therefore, the moles will be:

S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles

Now, the respective masses are:
S = 32 * 3.64 = 116.48 grams
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7 0
2 years ago
A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?
arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0
2 years ago
An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered
Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0
1 year ago
On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N2 in air is 0.78 atm and the concentration of
joja [24]

Answer : The partial pressure of nitrogen gas is, 2.94 atm

Explanation:

According top the Henry's Law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas.

C\propto P

C=K_H\times P

K_H is Henry's constant.

or,

\frac{C_1}{C_2}=\frac{P_1}{P_2}

where,

C_1 = initial concentration of gas = 5.3\times 10^{-4}M

C_2 = final concentration of gas = 2.0\times 10^{-3}M

P_1 = initial partial pressure of gas = 0.78 atm

P_2 = final partial pressure of gas = ?

Now put all the given values in the above formula, we get the final partial pressure of the gas.

\frac{5.3\times 10^{-4}M}{2.0\times 10^{-3}M}=\frac{0.78atm}{P_2}

P_2=2.94atm

Therefore, the partial pressure of nitrogen gas is, 2.94 atm

5 0
2 years ago
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