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hjlf
3 years ago
14

How many atoms are in 3.2 moles of chromium?

Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0

Hey there!

1 mole has 6.022 x 10²³ atoms.

Multiply that by 3.2 because we have 3.2 moles.

3.2 x 6.022 x 10²³

1.9 x 10²⁴

There are 1.9 x 10²⁴ atoms in 3.2 moles of chromium.

Hope this helps!

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irakobra [83]

Answer:

<u><em>Arrhenius Acid:</em></u>

According to Arrhenius concept, Acids are proton donors.

Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>

<u><em>Arrhenius Base:</em></u>

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>

<u><em></em></u>

6 0
3 years ago
Solutions and colloids which contain similarly sized particles can be effectively separated using paper filtration.
mojhsa [17]
False cause there particles arent large enough to be filtered
6 0
3 years ago
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Which of the following is a reasonable way of describing the density of a gas at STP
dedylja [7]

Answer:the answer is a

Explanation:

8 0
2 years ago
For an alloy that consists of 78.2 g copper, 103.5 g zinc, and 2.8 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c
TEA [102]

Answer:

Explanation:

Mass percent is defined as the mass of an element divided by the sum of masses of all the elements multiplied by 100. It is generally used to define the concentration. It does not depend on concentration.

It is given as;

Mass percent = (mass of an element / Total mass of the compound ) × 100

Mass of  compound (alloy) = 78.2 g copper  + 103.5 g zinc +2.8 g lead = 184.5 g

(a) Cu

Mass of Cu = 78.2

Mass percent = (78.2 / 184.5) * 100 = 42.38%

(a) Zn

Mass of Zn = 103.5

Mass percent = (103.5 / 184.5) * 100 = 56.10%

(c) Pb

Mass of Pb = 2.8

Mass percent = (2.8 / 184.5) * 100 = 1.52%

8 0
3 years ago
The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
aev [14]

Answer: <em>C₃H₆O</em>


Explanation:


1)  First, calculate the chemical composition, which you do using balance mass using the data given.


2) Data:


i) mass of ethyl butyrate: 2.22 mg

ii) mass of CO₂: 5.06 mg

iii) mass of H₂O: 2.06 mg


3) Solution


i) All the carbon in CO₂ produced was present in the ehtyl butyrate sample.


ii) The molar mass of CO₂ is 44.01 g/mol


iii) Use proportionality:


12.01 g C / 44.01 g CO₂ = x / 5.06 mg C  ⇒ x = 1.38 mg  C


iv) All the H in H₂O was present in the original sample of ethyl butyrate


v) The molar mass of H₂O is 18.015 g/mol


vi) Use proportionality


2.016 g H / 18.015 g H₂O = x / 2.06 mg H ⇒ x = 0.23 mg H


vii) Mass balance:


mass of O in the sample = mass of the sample - mass of H - mass of C


mass of O = 2.22 mg - 1.38 mg - 0.23 mg = 0.61 mg O.


viii) Composition:

C = 1.38 mg

H = 0.23 mg

O = 0.61 mg


Which in terms of composition is the same that:

C = 1.38 g

H = 0.23 g

O = 0.61 g


ix) Divide each by the molar atomic mass of the corresponding element, to obtain the number of moles:


C = 1.38 g / 12.01 g/mol = 0.115 mol

H = 0.23 g / 1.008 g/mol = 0.228 mol

O = 0.61 g / 16 g/mol = 0.0381 mol


x) Divide each by the least number of moles to obtain mole proportion


C = 0.115 / 0.0381 = 3.02 ≈ 3

H = 0.228 / 0.0381 = 1.98 ≈ 5.98 ≈ 6

O = 0.0381 / 0.0381 = 1


xi) Therefore the empirical formula searched is <em>C₃H₆O</em>.


7 0
3 years ago
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