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Kruka [31]
3 years ago
15

Is my answer correct?

Chemistry
1 answer:
nika2105 [10]3 years ago
4 0

Your answer is correct. Quantum numbers,n and l can only take positive values. 

Where:

n = cannot be zero

l = any number from zero to n-1

<span>lm = any number between –l and +l including zero</span>

<span>ms = either -1/2 or +1/2</span>

<span><span>
Looking at the choices, </span>(3, 2, 1, - 1/2 ) is the only choice that satisfy the criteria.</span>

<span>


</span>

 

<span> </span>

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Which of the following is NOT an important physical property of matter?
IgorLugansk [536]
The best option is melting point 
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What is an ion and how is it formed?
grin007 [14]

Answer: Ions are an atom or molecule with a net electric charge due to the loss or gain of one or more electrons. They are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells.

Explanation:

When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

6 0
3 years ago
a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


4 0
3 years ago
The students who conducted the shadow length experiment concluded that their results concluded that their results supported thei
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6 0
3 years ago
4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/
nalin [4]

Answer:

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.

Explanation:

Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.

Mass = Density × Volume

Mass of t-butyl alcohol = 0.79 g/mL\times 10.4 mL=8.219 g

Moles of t-butyl alcohol  =\frac{8.219 g}{74.12 g/mol}=0.11084 mol

Moles of 1,4-dimethoxybenzene = \frac{5.6 g}{138.17 g/mol}=0.04052 mol

According to reaction 2 mol of  t-butyl alcohol reacts  with 1 mol of 1,4-dimethoxybenzene.

Then 0.11084 moles of t-butyl alcohol will react with :

\frac{1}{2}\times 0.11084 mole=0.05542 mol of 1,4-dimethoxybenzene.

This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.

According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of  1,4-di-t-butyl-2,5-dimethoxybenzene.

Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:

\frac{1}{1}\times 0.04052 mol= 0.04052 mol of 1,4-di-t-butyl-2,5-dimethoxybenzene.

Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:

0.04052 mol × 250.37 g/mol = 10.144 g

Percentage yield:

\frac{Experimental}{Theoretical}\times 100

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:

Experimental yield = 4.2 g

Theoretical yield = 10.144 g

\frac{4.2 g}{10.144 g}\times 100=41.40\%

4 0
3 years ago
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