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Ierofanga [76]
3 years ago
7

Consider the total ionic equation below.

Chemistry
1 answer:
LenaWriter [7]3 years ago
7 0

Answer:

The spectator ions in this equation are :

CrO²⁻₄ and Ba²⁺

Explanation:

Spectator ions : These are the ions which are present in the solution but do not take part in chemical reaction.They are present as such in  both reactant as well as product .

In this reaction,

1.H+ and OH- are combined and converted into water (H2O) : Changed to H2O

So , H+ and OH- are not Spectator ions

2.CrO²⁻₄ and Ba²⁺ are unchanged . Present as such in product and reactants

CrO²⁻₄ and Ba²⁺are Spectator ions

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The volume of a fixed amount of gas is doubled, and the absolute temperature is doubled. According to the ideal gas law, how has
neonofarm [45]

Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

3 0
3 years ago
Why must the mass of magnesium be less than 0.09 g?
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The mass of magnesium should be less than 0.09g to enable a faster reaction rate. Magnesium reacts to form a white coating around it which stops the reaction. The lesser the gram the faster the reaction before the coating is formed. It is also advisable to use magnesium fillings to increase the rate of reaction.
7 0
3 years ago
If a piece of metal has a density of 9.865 g and a volume of 14 ml, what is its mass?
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Answer:

I think the answer is 138.11 gram

Explanation:

8 0
3 years ago
Which energy profile best shows that the enthalpy of formation of cs2 is 89.4 kj/mol?
Kruka [31]
Answer is: B. C(s) + 2S(s) + 89.4 kJ → CS2(l).

Missing question:
A. C(s) + 2S(s) → CS2(l) + 89.4 kJ.
B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
C. C(s) + 2S(s) + 89.4 kJ → CS2(l) + 89.4 kJ.
D. C(s) + 2S(s) → CS2(l).
Because enthalpy of the system is greater that zero, this is endothermic reaction (<span>chemical reaction that absorbs more energy than it releases)</span>, heat is included as a reactant.
5 0
3 years ago
If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac
STALIN [3.7K]

<u>Answer:</u> The theoretical yield of barium sulfate is 50.9 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol

  • <u>For barium chloride:</u>

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = \frac{1}{1}\times 0.228=0.228mol of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = \frac{1}{1}\times 0.228=0.228moles of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g

Hence, the theoretical yield of barium sulfate is 50.9 grams

7 0
3 years ago
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