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3 years ago

Which Element is found in the Noble Gases Group 8

Chemistry

1 answer:

Svetach [21]3 years ago

4 0

(Ar) Argon

Explanation:

The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

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Can someone check my chem questions?

alekssr [168]

All of them are correct! good!!

7 0

3 years ago

Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the

jek_recluse [69]

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

4 0

3 years ago

Read 2 more answers

What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC

faltersainse [42]

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

This makes Na to be unbalanced. Now to balance Na, put 2 in front of NaCl as illustrated below

2NaOH + CuCL2 —> 2NaCl + Cu(OH)2

Now the equation is balanced.

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

4 0

4 years ago

Which of the following is true of group 6A?

USPshnik [31]

Answer:

The answer to your question is: Includes sulfur and gain two electrons

Explanation:

Includes Chlorine This option is wrong, Chlorine belongs to group VII.

Includes Sulfur This option is true, Group VI includes Oxygen, Sulfur, Selenium, Tellurium.

Gain 2 electrons . This option is true, Elements in group VI have six valence electrons so they gain to electrons to become estable.

Tend to form +2 ions This option is wrong, this elements form -2 ions

Have 5 valence electrons This option is wrong, this elements have 6 valence electrons.

7 0

3 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago