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3 years ago

Which of he following compounds is always part of an aqueous solution

Chemistry

1 answer:

weqwewe [10]3 years ago

7 0

Answer: Water because if we say aqueous it means it resembles water or related to water.

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A 7.337 gram sample of chromium is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 9.595 g. Dete

Thepotemich [5.8K]

<h3>Answer:</h3>

Empirical formula is CrO

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of sample of Chromium as 7.337 gram
Mass of the metal oxide formed as 9.595 g

We are required to determine the empirical formula of the metal oxide.

<h3>Step 1 ; Determine the mass of oxygen used </h3>

Mass of oxygen = Mass of the metal oxide - mass of the metal

= 9.595 g - 7.337 g

= 2.258 g

<h3>Step 2: Determine the moles of chromium and oxygen</h3>

Moles of chromium metal

Molar mass of chromium = 51.996 g/mol

Moles of Chromium = 7.337 g ÷ 51.996 g/mol

= 0.141 moles

Moles of oxygen

Molar mass of oxygen = 16.0 g/mol

Moles of Oxygen = 2.258 g ÷ 16.0 g/mol

= 0.141 moles

<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>

Mole ratio of Chromium to Oxygen

Cr : O

0.141 mol : 0.141 mol

1 : 1

Empirical formula is the simplest whole number ratio of elements in a compound.

Thus the empirical formula of the metal oxide is CrO

6 0

3 years ago

g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0

3 years ago

Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentra

svetoff [14.1K]

Answer:

[Cl2] equilibrium = 0.0089 M

Explanation:

<u>Given:</u>

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

<u>To determine:</u>

The equilibrium concentration of Cl2

<u>Calculation:</u>

Set-up an ICE table for the given reaction:

$SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)$

I 0 0.0546 0.0546

C +x -x -x

E x (0.0546-x) (0.0546-x)

$Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M$

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M

5 0

3 years ago

Read 2 more answers

1. If I have 3 identical books that all have a length of 25cm, a

Ksenya-84 [330]

Answer:

12,375 cm³

Explanation:

v=l×w×h

Since there are 3 books the equation would then be

v=3(l×w×h)

v=3(25×15×11)

v=3(4,125)

v=12,375 cm³

5 0

3 years ago

A mixture of benzene and toluene has a total vapor pressure at 25°C of 45.06 mmHg. What is the partial pressure of benzene in th

SashulF [63]

Answer:

I just need points so ignore my answer and thanks

3 0

3 years ago