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2 years ago

What are the spectators ions in this reaction?

Chemistry

1 answer:

miv72 [106K]2 years ago

4 0

Answer:

Option C. Na⁺(aq) and Cl¯(aq)

Explanation:

From the question given above, we obtained the following:

Na₂SO₄(aq) + CaCl₂(aq) —> CaSO₄(s) + 2NaCl(aq)

Ionic Equation:

2Na⁺(aq) + SO₄²¯(aq) + Ca²⁺(aq) + 2Cl¯(aq) —> CaSO₄(s) + 2Na⁺(aq) + 2Cl¯(aq)

From the ionic equation above, we can see that Na⁺(aq) and Cl¯(aq) are present on both side of the equation.

Therefore, Na⁺(aq) and Cl¯(aq) are the spectator ions because they did not participate directly in the reaction.

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Example of a negative impact of acids in our everyday world

Marizza181 [45]

The acid clouds fog the important nutrients like leave and needles . this loss of nutrient make it easier for infections , insects , and cold weather to damage trees and forests

4 0

2 years ago

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +

aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

$\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles$

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

= -3608 kJ

≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

5 0

2 years ago

PLEASE HELP I got this wrong on an exam but got it wrong and i don’t know why!

lord [1]

Answer:

D.5,55cm+/-0,005cm

Explanation:

so this is the answer

5 0

2 years ago

The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval

worty [1.4K]

<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t

The average rate of disappearance of ozone ... is found to

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t

rate = -(1/2)*(-7.25 × 10^–3 atm)

= 3.625 × 10^–3 atm

Now use the other part of the expression:

rate = +(1/3)∆[O2)∆t

3.625 × 10–3 atm = +(1/3)∆[O2)/t

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

= 1.0875 x 10-2 atm over the same time interval

4 0

2 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

2 years ago