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3 years ago

How is laughing gas made first answer gets brainleist and 100 points

Chemistry

2 answers:

Ket [755]3 years ago

5 0

Answer:

Make nitrous oxide or laughing gas by heating ammonium nitrate and collecting the vapor by bubbling it up into a container over water. ... It's easy to make nitrous oxide or laughing gas at home or in the lab. All you need is a heat source and ammonium nitrate

Explanation:

aleksley [76]3 years ago

4 0

Answer:

laughing gas is made by heating ammonium nitrate and collecting the vapor by bubbling it up into a container over water

Explanation:

Nitrous Oxide is also called laughing gas or happy gas due to its intoxicating effects when inhaled. It was discovered around 1772 by the English scientist and clergyman Joseph Priestley 

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PLEASE HELPP!!!!

Natasha_Volkova [10]

Answer:

The combined law is:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

See the derivation below.

Explanation:

1. Boyle's law:

$PV=constant\\\\P_1V_1=P_2V_2$

2. Charles' law:

$\dfrac{V}{T}=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

3. Gay-Lussac’s law

$\dfrac{P}{T}=constant\\\\\\\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

4. Summary:

PV = K

V/T = K'

P/T = K''

Mulitply PV by V/T and P/T

$PV\times \dfrac{V}{T}\times \dfrac{P}{T}=K\times K'\times K''\\\\\\\dfrac{P^2V^2}{T^2}=constant\\\\\\\sqrt{\dfrac{P^2V^2}{T^2}}=\sqrt{constant}\\\\\\\dfrac{PV}{T}=constant$

Thus:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Which is the combined law.

4 0

4 years ago

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate

Anvisha [2.4K]

Answer:

$-3272$ kJ/mol

Explanation:

Given and known facts

Mass of Benzene $= 0.187$ grams

Mass of water $= 250$ grams

Standard heat capacity of water $= 4.18$ J/g∙°C

Change in temperature ΔT $= 7.48$ °C

Heat

$=250 * 4.18 * 7.48\\=7816.6 \\=7.82$

Heat released by benzine is - 7.82 kJ

Now, we know that

$0.187$ grams of benzene release $= -7.82$ kJ heat

So, $1$ g benzine releases

$\frac{ -7.82 }{0.187}\\= -41.8$

kJ/g

$0.187 * \frac{1}{78.108}=0.00239$ mol C6H6

Heat released

$= \frac{-7.82}{ 0.00239}$

$=-3272$ kJ/mol

4 0

3 years ago

What are the difference between an animal cell and plant cell

Bingel [31]

Answer:

ANIMAL CELL

Animal cells are generally small in size.

They don't have a cell wall

Except the protozoan Euglena, no animal cell possesses plastids.

Vacuoles in animal cells are many, small and temporary.

Animal cells have a single highly complex and prominent Golgi apparatus.

Animal cells have centrosome and centrioles.

PLANT CELL

Plant cells are larger than animal cells.

The plasma membrane of plant cells is surrounded by a rigid cell wall of cellulose.

They have Plastids

Most mature plant cells have a permanent and large central sap vacuole.

Plant cells have many simpler units of Golgi apparatus, called dictyosomes.

Plant cells lack centrosome and centrioles.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Which of the following is NOT correct about elements?

Lynna [10]

Answer: B. Elements are represented by chemical formulas.

Elements are pure substances, which means that they cannot be broken down into simpler substances. The element is the most basic substance that exists, breaking it down further means breaking it down into protons, neutrons, and electrons, which is no longer a substance.

Elements have chemical properties that allow them to form different types of bonds with other elements.

However, elements *alone* are not represented by their chemical formulas. Only chemical bonds or ions are represented by a chemical formula.

3 0

4 years ago

Read 2 more answers

. How much energy is lost by a 30.0g sample of water that decreases in temperature from 56.7C to 25.0C?

lbvjy [14]

Answer:

Q = -3980.9 j

Explanation:

Given data:

Mass of sample = 30 g

Initial temperature = 56.7 °C

Final temperature = 25 °C

Specific heat of water = 4.186 j/g.°C

Amount of heat released = ?

Formula:

Q = m.c.ΔT

Q = heat released

m = mass of sample

c = specific heat of given sample

ΔT = change in temperature

Solution:

ΔT = T2 -T1

ΔT = 25 °C - 56.7 °C = - 31.7°C

Q = m.c.ΔT

Q = 30 g × 4.186 j/g.°C × - 31.7°C

Q = -3980.9 j

8 0

3 years ago