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kramer
3 years ago
15

Of the well-characterized elements of group 7A, which would exhibit the following properties?

Chemistry
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer: it reacts with the alkali metals (M) to form a salt MX, where X is the halogen.

Explanation: Group 7A elements are halogen and they react with alkali metals like Sodium or potassium to form a salt like NaCl

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A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the
Kamila [148]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

 <u>   P ∝ T</u>

(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

   T_{2} =\frac{333*4}{1.8}

    T_{2} = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0
3 years ago
How many moles of CO are there in 1.40 x 10^24 molecules.
viva [34]

Answer:

Explanation:

Mole = no. Molecules/6.02×10^23

Mole = (1.40×10^24)/(6.02×10^23)

Mole = 2.33mole

6 0
3 years ago
Redox 1/2 reaction Cu(s) + 2 AgC2H3O2(aq) = Cu(C2H3O2)2(aq) + 2 Ag(s)
Anna11 [10]

Explanation:

 Reaction:

              Cu  +     2AgC₂H₃O₂     →    Cu(C₂H₃O₂)₂    +     2Ag

The problem is to split the reaction into oxidation and reduction halves:

 The oxidation half is the sub-reaction that undergoes oxidation

  The reduction half is the one that undergoes reduction:

The ionic equation:

 Cu   +  2Ag⁺   +   2C₂H₃O₂⁻   →   Cu²⁺  +  2C₂H₃O₂⁻   +  2Ag

Oxidation half:

     Cu →  Cu²⁺ + 2e⁻

 

Reduction half:

    2Ag⁺ + 2e⁻ → 2Ag

C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.

learn more:

Oxidation state brainly.com/question/10017129

#learnwithBrainly

4 0
3 years ago
553.5 mL ?:cm^3<br> how do i get my answer? and what is the answer?
dmitriy555 [2]

Answer:

553.5 cm3

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6 0
3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
2 years ago
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