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3 years ago

9

The chemical formula for the unknown molecule shown above is...

2 answers:

solmaris [256]3 years ago

8 0

Answer:

dsrthsrhrhrhrthrsh

Explanation:

notka56 [123]3 years ago

4 0

Answer:

the answer is A

Explanation:

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Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?

77julia77 [94]

N = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa, R = 8.314 L.kPa/K.mol

PV = nRT

V = nRT / P substituting.

V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (<span>101.325 kPa)

That is the answer, but it is not among the options you provided. Check your options properly.</span>

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3 years ago

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True or false? an element consists of only single atoms of that element. <br> a. True <br> b. False

larisa [96]

From what i read, the answer should be b--false

7 0

3 years ago

When the white part is on the left the moon is.??

Kazeer [188]

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3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28

ivanzaharov [21]

<u>Answer:</u> The mass of second isotope of indium is 114.904 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

.....(1)

Let the mass of isotope 2 of indium be 'x'

<u>For isotope 1:</u>

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

<u>For isotope 2:</u>

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

$114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu$

Hence, the mass of second isotope of indium is 114.904 amu

4 0

3 years ago