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Effectus [21]
3 years ago
7

Depending on the reaction conditions, elements X and Y react by a weight ratio of either 1 part of X and 1.982 parts of Y, or 1

part of X and 5.946 parts of Y, forming two different compounds. Which of the following pairs of substances is inconsistent with the above data?
(a) X3 Y and X3Y3
(b) X3Y2 and X3Y6
(c) XY and X3Y
(d) XY and X3Y3
(e) X3Y2 and XY2
Chemistry
1 answer:
Natalija [7]3 years ago
3 0

Answer:

(c) and (d)

Explanation:

The law of definite proportions states that elements in compounds always have a fixed atomic ratio independent of any conditions.

In the context of this problem, we only know the mass ratio of X to Y, but we don't know the molar ratio, so we cannot derive the empirical formula.

However, we notice that in the second compound, mass of Y increased by a fraction of:

\frac{5.946}{1.982} = 3

Since mass is directly proportional to moles, in the second compound we should have three times larger number of atoms of Y for the same number of atoms of X (1 part in each case lets us assume this).

a. Notice that X is kept constant and Y increases by a factor of 3, so it's consistent.

b. Similarly to (a), X is kept constant and Y increases by a factor of 3, so it's consistent.

c. Here, Y is kept constant. X increases by a factor of 3. This is inconsistent with the data, since if we start with XY, then we'd have XY_3 as our second pair, since we should have three times greater number of Y atoms in the second structure.

d. The molar ratio here in each pair is the same, XY is the empirical formula for X_3Y_3 = (XY)_3. This means it's inconsistent, as here both formulas have the same ratio of X to Y.

e. This is consistent, since for a constant Y, in the second structure we should have one third of X atoms to obtain the 3 : 1 ratio.

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what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require
vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of HCOO^{-}

So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of HCOO^{-} produced = \frac{29.80\times 0.3567}{1000}moles=0.01063moles

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of HCOO^{-} = \frac{0.01063\times 1000}{54.80}M=0.1940M

At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

5 0
3 years ago
What is the main difference between an isotope and an ion
eimsori [14]
An isotope is the vary of neutrons in an element, causing its atomic mass to change. While an ion is a charged atom that bonds to be stable. 
6 0
4 years ago
Carbon-12 and carbon-13 are isotopes of carbon.
ohaa [14]

Answer:

Isotopes of an element share the same number of protons but have different numbers of neutrons. Let's use carbon as an example. There are three isotopes of carbon found in nature – carbon-12, carbon-13, and carbon-14. All three have six protons, but their neutron numbers - 6, 7, and 8, respectively - all differ.

Explanation:

4 0
3 years ago
Read 2 more answers
How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?
nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

  • Concentration of the concentrated solution (C₁): 2 M
  • Volume of the concentrated solution (V₁): ?
  • Concentration of the diluted solution (C₂): 0.1 M
  • Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

4 0
3 years ago
Find a mole of 0.0960 g of H2SO4
nignag [31]
Molar mass H₂SO₄ = 98.079 g/mol

1 mol -------- 98.079 g
? mole ------ 0.0960 g

moles = 0.0960 * 1 / 98.079

= 0.0960 / 98.079

= 9.788 x 10⁻⁴ moles

hope this helps!
3 0
4 years ago
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