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3 years ago

How many moles are contained in 3.131 × 1024 particles?

Chemistry

2 answers:

KiRa [710]3 years ago

4 0

All are wrong, it would be 5.324e-21 moles

AleksAgata [21]3 years ago

3 0

<u>Given:</u>

The number of particles = 3.131 * 10²⁴

<u>To determine:</u>

The number of moles corresponding to the given particles

<u>Explanation:</u>

1 mole corresponds to Avogadro's number of particles = 6.023 * 10²³ particles

Therefore, the given particles would correspond to:

3.131* 10²⁴ particles * 1 moles/6.023 * 10²³ particles = 5.199 moles

Ans: A) 5.199 moles are contained in 3.131* 10²⁴ particles

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Which of the following is not part of Dalton's atomic theory?

lakkis [162]

I think its A because some scientist from the Department of Physics of Northeastern University found out that is not a part of Dalton's atomic theory.

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3 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

2. In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] =

notsponge [240]

Kc= concentration of product divided by concentration of reactant
NO + NO2 ----> N2O3

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Kc= ( 1.3 )/{ (3.9)(3.8) }

Kc=0.088 ( answer B)

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3 years ago

Read 2 more answers

SpongeBob & Friends and the Scientific Method

yarga [219]

Answer:

1. Which people are in the control group? The people who received the mint without the secret ingredient

(Group B) would be the control group.

2. What is the independent variable? Secret ingredient in the breath mint

3. What is the dependent variable? Amount of breath odor (or bad breath)

4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the

amount of breath odor more than half the time, but it is not 100% effective.

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3 years ago

Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones

Scilla [17]

Answer:

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Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

$^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is $_{97}^{345}\textrm{Bk}$

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3 years ago