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creativ13 [48]
3 years ago
14

(WILL GIVE BRAINLIEST IF YOU ANSWER ALL 4) Chemistry

Chemistry
1 answer:
Kitty [74]3 years ago
5 0

Answer:

See explanations

Explanation:

a. Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl

b. Freezing Pt Depression

     1. Sprinkling salt on icy surfaces

    2. Using antifreeze in automobile cooling systems

    3. <em>Not an application </em>

    4. Using salt to make ice cream

c. pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4

d. H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O  (proton donor)

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A water molecule is made of hydrogen and oxygen atoms and has the same formula is H2O how was the mixture of hydrogen oxygen dif
maksim [4K]

A mixture of hydrogen and oxygen would not need to have a certain ratio of hydrogen atoms to oxygen atoms to differ from the compound known as water.

<h3>Why is water a compound not a mixture?</h3>

A water molecule has 2 hydrogen atoms bonded chemically to oxygen atoms and has the formula H₂O.

In a chemical reaction, two atoms react together chemically to form a new product which means the reactants lose their individual properties and obtain new properties of the product formed.

Hence, water is a compound as hydrogen and oxygen atoms are bonded chemically to each other.

On the other hand, if oxygen and hydrogen form a mixture which is a physical change then they retain their individual properties in the mixture. They do not need to be in a specific ratio and there are no chemical changes taking place.

Chemical change determines if a substance is a mixture or a compound.

Thus, a mixture of hydrogen and oxygen is different from water if it doesn't have a certain ratio of hydrogen atoms to oxygen atoms.

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4 0
2 years ago
A particle with charge +1 nC (a nanocoulomb is 1 10-9 C) is located at the origin. What is the electric field due to this partic
Zolol [24]

Answer:

E=35.97\frac{V}{m}

Explanation:

The electric field of a point charge can be calculated using the next equation:

E=\frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}

Where:

\epsilon_0=Absolute\hspace{3}permittivity\hspace{3}of\hspace{3}free\hspace{3}space\approx 8.85\times 10^{-12} \\Q=Charge\hspace{3}magnitude\\r=Distance

The distance can be calculated as:

r=\sqrt{0.5^2+0^2+0^2} =0.5

Using the data provided by the problem:

E=\frac{1}{4\pi (8.85\times 10^{-12}) } \frac{1\times 10^{-9} }{0.5^2} =35.96721878\approx35.97V/m

8 0
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Mass of oxygen required = 19.5 moles × 16 g/mol = 312 g.

Explanation:

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Answer:

The second option

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