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3 years ago

What do scientists use to answer scientific questions?

Chemistry

1 answer:

VladimirAG [237]3 years ago

7 0

Answer: The scientific method

Explanation:

A P E X

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For a principal quantum number n, how many atomic orbitals are possible?

slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

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3 years ago

A sample of helium gas is in a sealed rigid container. What occurs as the temperature of the sample is increased

hammer [34]

The electrons will move more rapidly resulting in a higher pressure even at a consistent volume

3 0

3 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

3 years ago

An isotope of cesium has 55 protons and 78 neutrons. What is the mass number of this isotope?

mixer [17]

The only stable caesium isotope is 133Cs, with 78 neutrons.

Hope this answers your Q

4 0

3 years ago

Question 3 0

Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0

3 years ago