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3 years ago

Find the mass of 3.00 mol of acetic acid, C2H4O2.

1 answer:

5 0

Molar mass

C₂H₄O₂ = 60.0 g/mol

n = mass / molar mass

3.00 = mass / 60.0

m = 3.00 * 60.0

m = 180 g of <span>C₂H₄O₂

hope this helps!</span>

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CAN SOMEONE HELP ME?!?! IM CONFUSED

Sati [7]

Answer: Solar winds have an influence all the way to about 160 AU from the Sun. About how many kilometers is that?

24,000,000,000 km

48,000,000,000 km

240,000,000,000 km

480,000,000,000 km

Explanation:

4 0

3 years ago

1. Find the number of protons, neutrons, electrons and electronic structure of flourine 919F.

77julia77 [94]

1. ₉¹⁹F

2. Mg(OH)₂

3. 2H₂ + O₂ ⇒ 2H₂O

<h3>Further explanation</h3>

1. Fluorine, atomic number : 9 , mass number = 19

Symbol : ₉¹⁹F

protons=electrons=atomic number = 9

neutrons = mass number - atomic mass

$\tt n=19-9=10$

Configuration : [He] 2s² 2p⁵

2. Magnesium hydroxide is an ionic compound and is a strong base consisting of 2 ions:

Positive ion: Magnesium: Mg²⁺

negative ion: Hydroxide: OH⁻

The charges of the two are crossed, so that the compound becomes:

Mg(OH)₂

3. Reaction :

H₂ + O₂ --------> H₂O

give coefficient :

aH₂ + bO₂ --------> H₂O

H, left = 2a, right 2⇒2a=2⇒a=1

O, left = 2b, right 1⇒2b=1⇒b=0.5

Reaction becomes :

H₂ + 0.5O₂ --------> H₂O x 2

2H₂ + O₂ --------> 2H₂O

4 0

3 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago

Which of the following shows 13,400 as having 4 significant figures?

Gnom [1K]

A would be your answer

5 0

3 years ago

Answer ASAP 25 points

kykrilka [37]

<u>Answer:</u> The correct answer is $NH_3$ is the base and $NH_4Cl$ is the salt.

<u>Explanation:</u>

Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce salt and water as products.

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and a base is defined as a substance which accepts protons.

For the given chemical reaction:

$NH_3+HCl\rightarrow NH_4Cl$

Here, $NH_3$ is a base because it is accepting proton and HCl is an acid because it is loosing proton.

$NH_4Cl$ is considered as a salt because it is formed by the combination of acid and base.

Hence, the correct answer is $NH_3$ is the base and $NH_4Cl$ is the salt.

8 0

3 years ago

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