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Vladimir [108]
2 years ago
11

1. The electron configuration below represents the ground state of an atom.

Chemistry
1 answer:
Natalka [10]2 years ago
4 0

Answer:

{S}^{16}

group 16 period 2 of the periodic table

note: that is not the electronic configuration, that is the Bohr model.

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12. A beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. Will it float in water,
Greeley [361]

Taking into account the definition of density and Archimedes' principle, the beginner bowling ball will float on the water.

But first it is neccesary to know that density is a quantity referred to the amount of mass in a certain volume of a substance or a solid object.

In other words, the density is the relationship between the weight (mass) of a substance and the volume that the same substance occupies.

The expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

In this case, a beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. This is:

  • mass= 4.9 kg= 4900 g (being 1 kg= 1000 kg)
  • volume= 5.4 L= 5400 mL (being 1L=1000 mL)

Replacing in the definition of density:

density=\frac{4900 g}{5400 mL}

Solving:

<u><em>density=0.907 </em></u>\frac{g}{mL}<u><em /></u>

On the other hand, Archimedes' principle says that an object immersed in a liquid experiences an upward vertical force equal to the weight of the volume of the dislodged liquid.

The sinking or floating of an object is determined by its density with respect to that of the liquid in which it is submerged.

Considering water as the liquid where the object is submerged in this case, an object with a higher density than water will sink. In contrast, an object with a lower density than water will float.

In this case, considering that water has a density of 1 \frac{g}{mL}, the bowling ball for beginners has a lower density. This indicates that, having a lower density than water, the object will float.

In summary, the beginner bowling ball will float on the water.

Learn more about density:

  • <u>brainly.com/question/952755?referrer=searchResults </u>
  • <u>brainly.com/question/1462554?referrer=searchResults</u>

5 0
3 years ago
JOSE BOUGHT 750 BAGS OF PEANUTS FOR 375 HE INTENDS TO SELL EACH BAG FOR 0.15 MORE THAN HE PAID HOW MUCH SHOULD HE CHARGE FOR EAC
labwork [276]
He paid 2 dollars for each bag. Then add 15 cents for that and he would charge $2.15 for each bag.
7 0
3 years ago
Read 2 more answers
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<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 NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
An amide may be produced by reacting an acid chloride with
Elanso [62]

Answer:

An amide may be produced by reacting an acid chloride with ammonia.

8 0
3 years ago
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Why didn't you just clean the spill with water?
Allisa [31]

Answer:

if you did it would probably make it bigger...

Explanation:

:)

3 0
3 years ago
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