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Kruka [31]
2 years ago
4

Compound A and Compound B are binary compounds containing only elements X and Y. Compound A contains 1.000 g of X for every 2.10

0 g of Y. Which mass ratio for Compound B below follows the law of multiple proportions with Compound A?a. 1.000 g X: 0.1621 g Y b. 1.000 g X: 0.7391 g Y c. 1.000 g X: 0.2579 g Y d. 1.000 g X: 0.2376 g Y e. 1.000 g X: 0.2733 g Y
Chemistry
1 answer:
kvasek [131]2 years ago
3 0

Answer:

a

Explanation:

The Law of Multiple Proportions is the third postulate of Dalton, and it stated that in a compound, the mass of one element combines with a fixed mass of the other element in a proportional ratio of whole numbers.

In the compound A, the mass of Y per gram of X is 2.100 g of Y per g of X. Thus, let's analyze the letters, and identify in which the ratio will be a whole number:

a. 0.1621 g of Y per g of X

ratio = 2.100 : 0.1621 (Dividing both for 0.1621)

ratio = 12.9 : 1

Which ca be approximate to 13 : 1, a whole number

b. 0.7391 g of Y per g of X

ratio = 2.100 : 0.7391 (Diving both for 0.7391)

ratio = 2.84 : 1

Which is not a whole number

c. 0.2579 of Y per g of X

ratio = 2.100 : 0.2579 (Dividing both for 0.2579)

ratio = 8.143 : 1

Which is not a whole number

d. 0.2376 g of Y per g of X

ratio = 2.100 : 0.2376 (Dividing both for 0.2376)

ratio = 8.384 : 1

Which is not a whole number

e. 0.2733 g of Y per g of X

ratio = 2.100 : 0.2733 (Dividing both for 0.2733)

ratio = 7.684 : 1

Which is not a whole number

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In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
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The volume of the container is 59.112 L

Explanation:

Given that,

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Read 2 more answers
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
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Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

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2 years ago
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