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3 years ago

the most common isotope of chromium has a mass number of 52. which notation represent a different isotope of chromium?

1 answer:

4 0

Not quite sure what your question is asking. Chromium-52, the most common isotope. Among lesser common isotopes there could be Chromium-53, Chromium-56, etc. I would have to look specifically at what exists. The percent abundance in the universe multiplied by each of these mass numbers and added determines the atomic mass number a displayed on the periodic table.

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What happens to particles during a physical change?

AnnZ [28]

Answer:

im pretty sure they heat up

Explanation:

as they use energy to go from solid to liquid to gas, they need heat to do so as they have a melting and boiling point

3 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:2 SO2(g) + O2(g) ⇌ 2 SO3(g). If initial concentrations are

oee [108]

This is an incomplete question, here is a complete question.

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

If initial concentrations are [SO₂] = 1.20 M, [O₂] = 0.45 M, and [SO₃] = 1.80 M, the system is

A) at equilibrium.

B) not at equilibrium and will remain in an unequilibrated state.

C) not at equilibrium and will shift to the left to achieve an equilibrium state.

D) not at equilibrium and will shift to the right to achieve an equilibrium state.

Answer : The correct option is, (A) at equilibrium.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The expression for reaction quotient will be :

$Q=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.80)^2}{(1.20)^2\times (0.45)}=5.0$

The given equilibrium constant value is, $K_c=5.00$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

Hence, the correct option is, (A) at equilibrium.

7 0

3 years ago

What is the water table?

horsena [70]

I think the answer is D. the upper surface of undergound water.

6 0

4 years ago

Read 2 more answers

A reaction will be spontaneous if the change in Gibbs free energy ____

aleksklad [387]

Answer:

C. is less than 0

Explanation:

The Gibbs free energy (G) of a system is defined as the energy that is used for the work done. The change in Gibbs free energy during a reaction shows the spontaneity and energetics of the reactions.

A reaction will be spontaneous if the change in Gibbs free energy will be negative or less than 0. Reactions with positive Gibbs free energy are non-spontaneous.

Hence, the correct answer is "C. is less than 0".

5 0

3 years ago