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ad-work [718]
3 years ago
6

Why is there a large difference between day and night temperatures in a desert?

Chemistry
1 answer:
liraira [26]3 years ago
6 0

Answer: Heat escapes rapidly at night because the air is so dry.

Explanation: The desert has little humidity and rarely is cloudy. These two factors allow the desert to heat up rapidly during the day and cool off quickly at night.

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Explain Garri industry<br>​
olasank [31]

Answer:

In West Africa, garri refers to the creamy granular flour obtained by processing the starchy tuberous roots of freshly harvested cassava. ... Flour foodstuffs mixed with cold or boiled water are a major part of the diet amongst the various ethnicities of Nigeria, Benin Republic, Togo, Ghana, Guinea, Cameroon and Liberia.

Explanation:

4 0
3 years ago
Container has 28% hydrogen gas and 72%of helium gas at 760 torr pressure. Calculate the partial pressure of helium gas ?
frosja888 [35]

uuuuuuuuuuuuuuuuuuuuuuum i think it the way its shaped

8 0
3 years ago
Which profile best describes the reaction C(s) + 2H2(g) → CH4(g),<br> AH = -74.9 kJ?
katrin [286]

Answer:

Option B. A

Explanation:

From the question given above, the following data were obtained:

C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ

From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.

For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change

(ΔH) is always negative.

Thus, diagram A (i.e option B) gives the correct answer to the question.

8 0
3 years ago
Read 2 more answers
Boyle's Law for gases is:
larisa86 [58]

Answer:

b. just a model and therefore accurate for no real gases

Explanation:

Boyle´s Law was determined and is applied for ideal gases, this is, those gases that:

- Do not have any interaction between their particles (neither attraction nor repulsion)

- Any collision between its particles is perfectly elastic

With these conditions, Boyle found that pressure and volume (in a constant temperature) are inversely proportional, which can be expressed as:

PV = k, where “k” is a constant

So, when pressure increases, volume decreases, and viceversa.  

If we have to different conditions (1 and 2) of pressure and volume (at constant temperature), this can be expressed like:

P₁V₁ = P₂V₂ = constant

The current and complete equation that links temperature, pressure, mass (in moles) and volume is:

PV = nRT

Real gases do not strictly comply with this law, as its particles has interactions and collisions are not perfectly elastic. This law is more accurate for gases with low molecular mass, and with low pressure and/or high temperature conditions (under these conditions, interactions can be neglected)

Another term that can gives us an idea whether a gas is ideal or not, is the compressibility coefficient Z:

Z = PV/RT

For ideal gases, Z = 1 , as long as the gas moves away from ideality, Z is totally different from 1

3 0
3 years ago
6) What types of atomic orbitals are in the third principal energy
Bumek [7]

Answer:

Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>

Explanation:

  • <u>OPTION A-:  s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
  • <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .

  • <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
  • <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .

<u>Thus , the correct option is C (s , p and d only .)</u>

6 0
4 years ago
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