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3 years ago

Which of these metals would you expect to be found in nature as a deposit of relatively pure element?

1 answer:

7 0

1. Which of these metals would you expect to be found in nature as a deposit of relatively pure element? The Answer is: Lithium. And question 2 is: Potassium - furthest away from lead in the reactivity series.

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

3 years ago

Calculate the volume of a 0.5M solution containing 20g of NaOH

Tju [1.3M]

Answer:

Explanation:

First, let us calculate the number of mole present in 20g of NaOH. This is illustrated below:

Mass = 20g

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of NaOH = 20/40 = 0.5mol

From the question given, we obtained the following data:

Molarity = 0.5M

Mole = 0.5mole

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.5/0.5

Volume = 1L

6 0

3 years ago

Read 2 more answers

C3H8 + 5 O2 → 3 CO2 + 4 H2O What is the number of atoms on each side of the equation?

Roman55 [17]

Answer: 1.2642*10²⁵ on both sides

Explanation:

First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides

There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21

Now use Avogrado's constant

21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³

= 1.2642*10²⁵

6 0

2 years ago

Read 2 more answers

Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O

Bess [88]

40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O

5 0

3 years ago

What is the name for a nuclear particle that has about the same mass as a neutron, but with a positive charge?

KATRIN_1 [288]

proton is the name for a nuclear particle that has about the same mass as a neutron, but with a positive charge.

3 0

3 years ago