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3 years ago

Half equation for oxygen to oxide ions, aluminium ions to aluminium, magnesium to magnesium ions

Chemistry

1 answer:

soldi70 [24.7K]3 years ago

5 0

2mg(s)+02(g)=2mg0 that was the correct answer

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Vrite skeleton equations for the following word equations.

almond37 [142]

Answer:

Explanation:

4 0

3 years ago

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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

A sample of water at 20 degrees c contains what bonds

vladimir2022 [97]

Covalent and hydrogen bonds

3 0

3 years ago

Large crystals are made from the slow cooling of magma. <br> a. True<br> b. False

Step2247 [10]

I would have to say the answer is a. True.

5 0

4 years ago

Read 2 more answers

(LO 37,45) A reaction is 75% complete in 45.0 min. How long after its start will the reaction be 50% complete if it is (A) first

seropon [69]

Answer:

A) t = 22.5 min and B) t = 29.94 min

Explanation:

Initial concentration, [A]₀ = 100

Final concentration = 100 -75 = 25

Time = 45 min

A) First order reaction

ln[A] − ln[A]₀ = −kt

Solving for k;

ln[25] − ln[100] = - 45k

-1.386 = -45k

k = 0.0308 min-1

How long after its start will the reaction be 50% complete?

Initial concentration, [A]₀ = 100

Final concentration, [A] = 100 -50 = 50

Time = ?

ln[A] − ln[A]₀ = −kt

Solving for k;

ln[50] − ln[100] = - 0.0308 * t

-0.693 = -0.0308 * t

t = 22.5 min

B) Zero Order

[A] = [A]₀ − kt

Using the values from the initial reaction and solving for k, we have;

25 = 100 - k(45)

-75 = -45k

k = 1.67 M min-1

How long after its start will the reaction be 50% complete?

Initial concentration, [A]₀ = 100

Final concentration, [A] = 100 -50 = 50

Time = ?

[A] = [A]₀ − kt

50 = 100 - (1.67)t

-50 = - 1.67t

t = 29.94 min

3 0

3 years ago