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Morgarella [4.7K]
2 years ago
12

16. These conditions occur over the course of a day. Cumulus clouds appear and create rain showers. The sky clears soon after th

e showers. The temperature drops after the sky clears. Which weather event is most likely moving through the area?
A. tornado

b. hurricane

c.cold front

d. warm front​
Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
6 0

Answer:

A

Explanation:

I believe it's A gggggg%gggggtggg

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Calculate the pH of a buffer that is 0.020 M HF and 0.040 M NaF. The Ka for HF is 3.5 × 10-4.
zzz [600]
 <span>pH = pKa + log ([base]/[acid]) = -log (3.5 x 10^-4) + log (0.040/.020) = 3.46 - 0.30 = 3.16</span>
4 0
3 years ago
Read 2 more answers
K for enolization of 2,4-cyclohexadienone is about 1013. Explain why the enol is so much more stable than the keto tautomer.
torisob [31]

Answer:

Because the value of K is huge.

Explanation:

The tautomer is a kind of isomer in which exist an equilibrium between a ketone and an enol, or between an aldehyde and an enol. So, in the enolization, the ketone is the reactant and the enol is the product.

The equilibrium reaction can be characterized by an equilibrium constant, which is the ratio of the concentration of the products by the concentration of the reactants.

Because the constant K is extremely large (10¹³) we can conclude that the concentration of the product will be greater than the concentration of the reactant, in the equilibrium. It means that the concentration of the enol will be greater.

So, the ketone is unstable and forms in a great amount the more stable product, the enol.

4 0
3 years ago
I need help click the image please help
Viefleur [7K]

Answer:

Answer:

a plant cell

Explanation:

because the chrolopast is green

Explanation:

6 0
2 years ago
1. Use your knowledge of valence electrons and how they affect bonds to figure out how carbon (Group 14) and oxygen (Group 16) w
erik [133]

From the fact that oxygen is in group 16 and carbon is in group 14, the structure of CO2 must be O=C=O. In methane, there is no bond between any of the hydrogen atoms. The structure of H2O2 is H–O–O–H.

Carbon is in group 14 hence it has four valence electrons and oxygen is in group 16 hence it has six valence electrons. This implies that each oxygen atom will share four electrons with carbon in a covalent bond to form the structure O=C=O.

In CH4, we know that carbon is tetravalent so it forms for bonds. Therefore, there is no bond between hydrogen atoms so it bonds with each hydrogen atom; hydrogen only forms one bond.

In H2O2, there is the peroxide ion that has the structure O-O. Hence, the correct structure of H2O2 is H–O–O–H.

Learn more: brainly.com/question/24775418

5 0
2 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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