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3 years ago

3. Give three examples of a pure substances

Chemistry

2 answers:

liq [111]3 years ago

7 0

Tin, sulfur, and water

Nadya [2.5K]3 years ago

5 0

Answer:

There will be weight, there will be volume, there will be height

Explanation:

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A chef is serving a lunch special with 1 sandwich and 2 corn cobs per plate. If there are 5 sandwiches and 14 corn cobs, what is

shtirl [24]

5 plates is the highest amount that can be served

There’s only 5 sandwiches so 7 is automatically ruled out, there’s 14 corn cobs and 5 sandwiches only need 10 so it works out

7 0

3 years ago

Read 2 more answers

Why are chemists interested in studying thermochemistry? a. Heat is often absorbed during reactions.b. Heat is often released du

Wittaler [7]

Answer:

a. Heat is often absorbed during reactions.

c. Heat is often released during chemical reactions.

Explanation:

Thermochemistry -

The study of thermochemistry involve the change in the amount of heat , during any physical or chemical process , is referred to as thermodynamics .

The focus of thermochemistry is on the changing amount of energy in the form of heat , on the system with respect to the surroundings .

The process like boiling , melting , sublimation , may require energy or releases energy , and hence are studied under thermochemistry .

Hence , from the given question ,

The correct options are - a , c.

6 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol$

For carbon dioxide gas:

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol$

The given chemical reaction follows:

$2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = $\frac{1}{2}\times 7994.12=3997.06mol$ of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia $(NH_3)$

5 0

3 years ago

HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

Answer: The correct statement is low temperature only, because entropy decreases during freezing.

Explanation:

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.

For a reaction to be spontaneous, $\Delta G$ must be negative.

$-ve=-ve-[T(-ve)]\\\\-ve=-ve+T$

From above equations, it is visible that $\Delta G$ will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

8 0

3 years ago

The figure shows different possible transitions of electrons as they move from higher energy states to lower energy states. Whic

Serjik [45]

I don’t know please answer for me

6 0

3 years ago