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mixer [17]
3 years ago
11

Which of the following is not a subatomic particle?

Chemistry
1 answer:
stepladder [879]3 years ago
7 0
<h3>Answer: C) Nucleus</h3>

The nucleus is not a subatomic particle, but rather made of the subatomic particles of protons and neutrons. An electron is also a subatomic particle, but it is not found in the nucleus. There are many other subatomic particles, but your teacher probably won't go over those terms until later.

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4. What is the total mass of ice that can be vaporized by 2100 kJ of heat energy?
anygoal [31]

Answer:

5 g

Explanation:

The heat required to vaporize ice is the sum of

i) Heat required to melt ice at 0°C

ii) Heat required to raise the temperature from  0°C to 100°C

iii) Heat required to vaporize water at 100°C

Thus;

H = nLfus + ncθ + nLvap

H= n(Lfus + cθ + Lvap)

Lfus = 6.01 kJ/mol

Lvap = 41 kJ/mol

c = 75.38

n =?

2100 = n(6.01  + 75.38(100) + 41)

n = 2100 KJ/7585.01 kJ/mol

n = 0.277 moles

Mass of water = number of moles * molar mass

Mass of water = 0.277 moles * 18 g/mol

Mass of water =  5 g

3 0
3 years ago
How many molecules are in 3 moles of potassium bromide (KBr)
sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

KBr

, as a conversion factor to help you find the mass of three moles of this compound.

So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

K

+

, and bromide anions,

Br

−

. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.

Use the periodic table to find the molar masses of these two elements. You will find

For K:

M

M

=

39.0963 g mol

−

1

For Br:

M

M

=

79.904 g mol

−

1

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

M

M KBr

=

39.0963 g mol

−

1

+

79.904 g mol

−

1

≈

119 g mol

−

So, if one mole of potassium bromide has a mas of

119 g

m it follows that three moles will have a mass of

3

moles KBr

⋅

molar mass of KBr



119 g

1

mole KBr

=

357 g

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

360 g

a

a

∣

∣

−−−−−−−−−

Explanation:

<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>

6 0
3 years ago
Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
According to boyle’s law, when the pressure on a gas in an enclosed container increases, the volume of that gas
Doss [256]
Decreases. Hope I helped
3 0
3 years ago
Calculate the mass in grams of benzophenone required to make a solution of 2.5 mmoles
kirill115 [55]

Answer:

0.46 grams (C₆H₅)₂CO

Explanation:

To find the mass of benzophenone ((C₆H₅)₂CO), you need to (1) convert mmoles to moles and then (2) convert moles to grams (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the sig figs of the given value (2.5 mmoles).

Molar Mass ((C₆H₅)₂CO): 13(12.011 g/mol) + 10(1.008 g/mol) + 15.998 g/mol

Molar Mass ((C₆H₅)₂CO): 182.221 g/mol

2.5 mmoles (C₆H₅)₂CO               1 mole                  182.221 g
-----------------------------------  x  ------------------------  x  -------------------  =
                                                1,000 mmoles            1 mole

=  0.46 grams (C₆H₅)₂CO

4 0
1 year ago
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