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2 years ago

How many mg does a 643 kg sample contain?

Chemistry

1 answer:

MariettaO [177]2 years ago

7 0

Answer:

$6.43x19^8mg$

Explanation:

Hello,

In this case, this unit conversion is performed by knowing that 1 kg of mass contains 1000 g and 1 g contains 1000 mg, therefore, the result is:

$643kg*\frac{1000g}{1kg} *\frac{1000mg}{1g}\\ \\=6.43x10^8mg$

Regards.

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Which industrial facility is most likely to produce toluene and benzene wastes?

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3. Scott and James work at a grocery store. After the grocery store closed, they were playing a game with a shopping cart and Sc

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2 years ago

Write the expression for the equilibrium constant Kp for the following reaction.Enclose pressures in parentheses and do NOT writ

maxonik [38]

Answer: The expression for $K_p$ is written below.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{P_{C}^c\times P_{D}^d}{P_{A}^a\times P_{B}^b}$

The partial pressure for solids and liquids are taken as 1.

For the given chemical equation:

$NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)$

The expression for $K_p$ for the following equation is:

$K_p=\frac{(P NH_3)\times (P H_2S)}{(P NH_4HS)}$

The partial pressure of $NH_4HS$ will be 1 because it is solid.

So, the expression for $K_p$ now becomes:

$K_p=\frac{(P NH_3)\times (P H_2S)}{1}$

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2 years ago

An atom of the element zinc has an atomic number of 30 and a mass number of 65. How many protons does an uncharged zinc atom hav

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Answer : The correct option is, 30 protons

Explanation :

Element = Zinc

Atomic number = 30

Atomic mass number = 65

As we know that the atomic number is equal to the number of electrons and number of protons.

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Number of neutrons = Atomic mass - Number of protons = 65 - 30 = 35

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2 years ago

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In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

$E_{cell} = E_{red} - E_{ox}$

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

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n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

ΔG° = -118x10³ J/mol

I hope it helps!

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2 years ago