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2 years ago

C= ( F - 32) / 1.8 86 F = ?

1 answer:

7 0

Hence, 84°F is 30°C in Celsius..

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Lost one extensive and one intensive property of a marker ink

emmainna [20.7K]

An extensive property is a property that deals with the amount of a substance. An extensive property of marker ink would be the volume of the ink in the pen. An intensive property is a property about the type of substance you're looking at. So, an intensive property of marker ink would be the color of the ink.

8 0

3 years ago

Give an example of how each compound is used or found in nature. Be sure to use complete sentences.

puteri [66]

Water is found in lakes and rivers. Its purpose in nature is to hydrate animals

Hydrogen gas is not abundant in nature, but hydrogen is abundant in water. Hydrogen (bonds) help ice float. Otherwise, freezing lakes would kill animals (it actually doesn't since the ice creates a "coat" above the water").

Ammonia is nitrogen-rich molecules that plant uses to get their nitrogen. It comes from the part of the nitrogen cycle where dead plants and animals are decayed.

Carbon dioxide is what we exhale. Plants "inhale" carbon dioxide and use that for photosynthesis.

Hydrogen sulfide is emitted by volcanoes and by anaerobic (oxygen-less) decay from bacteria.

Have an awesome day! I hope this helps.

8 0

3 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago

How much energy is required to raise the temperature of a 300.0

Montano1993 [528]

Answer:

Q = 1455.12 Joules.

Explanation:

Given the following data;

Mass = 300 grams

Initial temperature = 22.3

Final temperature = 59.9°C

Specific heat capacity = 0.129 J/gºC.

To find the quantity of energy;

$Q = mcdt$

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 59.9 - 22.3

dt = 37.6°C

Substituting the values into the equation, we have;

$Q = 300*0.129*37.6$

Q = 1455.12 Joules.

8 0

3 years ago

When performing labratory experiments, which of the following items is ALWAYS necessary to use?

densk [106]

All of the the answers are are correct but a fume hood is more for if you are dealing with chemicals that can produce fumes the are deadly to people

3 0

4 years ago