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2 years ago

11. What is another name for the transition metals? Group B elements noble gases Group A elements Group C elements

Chemistry

1 answer:

finlep [7]2 years ago

3 0

Answer:

Group B elements

Explanation:

Group B metals are referred to as transition metals. They are located in the middle of the periodic table between Group IIA and Group IIIA. They provide transition between the "base formers" on the left and "acid formers" on the right.

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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.

patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of $NH_{3}$ is as follows:.

$2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)$

And, initially only $NH_{3}$ is present.

The given data is as follows.

$P_{NH_{3}}$ = 2.3 atm at equilibrium

$P_{H_{2}} = 3 \times P_{N_{2}}$ = 0.69 atm

Therefore,

$P_{N_{2}} = \frac{0.69 atm}{3}$

= 0.23 aatm

So, $P_{NH_{3}}$ = 2.3 - 2(0.23)

= 1.84 atm

Now, expression for $K_{p}$ will be as follows.

$K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}$

$K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}$

= $\frac{0.23 \times 0.33}{3.39}$

= 0.0224

or, $K_{p} = 2.2 \times 10^{-2}$

Thus, we can conclude that the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is $2.2 \times 10^{-2}$ .

6 0

3 years ago

50 ml of 0.2 M acetic acid is shaken with 10 g activated charcoal. The concentration of acetic acid is reduced to 0.5 times the

Olegator [25]

Answer:

$0.03 g_{acid}/g_{charcoal}$

Explanation:

The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.

So, we need to calculate the mass of the acetic acid as follows:

$m = n*M = C*V*M$

Where:

n: is the number of moles = C*V

M: is the molecular mass = 60.052 g/mol

C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L

V: is the volume = 50 ml = 0.050 L

$m_{acid} = C*V*M = 0.10 mol/L*0.050 L*60.052 g/mol = 0.30 g$

Now, the amount of solute adsorbed per gram of the adsorbent is:

$\frac{m_{acid}}{m_{charcoal}} = \frac{0.30 g}{10 g} = 0.03 g_{acid}/g_{charcoal}$

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.

I hope it helps you!

3 0

2 years ago

. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.

asambeis [7]

Answer:

a=theory

b=law

c=hypothesis

Explanation:

because I know

6 0

2 years ago

Discuss why people use the scientific method

Lubov Fominskaja [6]

Answer:

They use the scientific method to study different aspects of how people think and behave. They also use the scientific method because it is evidence-based, standardized and objective in conducting experiments.

5 0

1 year ago

A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added and the

Levart [38]

Answer:

a. 102.6g/mol

b. Rb⁺

Explanation:

a. The unknown base, XOH, <em>Where X is the alkali metal</em>, reacts with HCl thus:

XOH + HCl → H₂O + XCl

<em>Where 1 mole of the hydroxide reacts per mole of acid. </em>

Moles of HCl spent to reach the equivalence point are:

17.0mL = 0.017L ₓ (2.50mol / L) = 0.0425moles of HCl

As 1 mole of HCl reacts with 1 mole of base, moles of base are 0.0425 moles.

Molar mass of a compound is the ratio between mass in grams and moles of substance, that is:

4.36g / 0.0425mol =

b) Molar mass of OH is 17g/mol, thus, the molar mass of the alkali metal is:

102.6g/mol - 17g/mol = 85.6g/mol

The mass of Rb is 85.5g/mol, thus, the identity of the alkali metal is

3 0

2 years ago