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levacccp [35]
3 years ago
6

How does the type of precipitation change currently?

Chemistry
1 answer:
Inga [223]3 years ago
5 0

Answer:

There are a couple different types of precipitation, though it is generally taken to mean a synonym for rain. It includes rain, snow, sleet, and hail. With the increase in global warming, there is more precipitation. In general, the type of precipitation varies with the season; we see less in the summer of all types, more snow, sleet, and hail in the winter, and more rain in the temperate seasons like fall.

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Thermochemical equations are chemical equations that include the __________ of the reaction.
irga5000 [103]

A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH. In variable form, a thermochemical equation would look like this:

A + B → CΔH = (±) #

Where {A, B, C} are the usual agents of a chemical equation with coefficients and “(±) #” is a positive or negative numerical value, usually with units of kJ.


please mark as brainliest

5 0
3 years ago
What happens to the glucose molecule during the process of cellular respiration? (5 points)
Dmitry [639]

The correct answer is option a, that is, it gets broken down.  

A set of metabolic reactions and procedures, which occurs in the cells of organisms to transform biochemical energy from nutrients into ATP, and then discharge waste components is known as cellular respiration. At the time of cellular respiration, a molecule of glucose gets dissociated slowly into water and carbon dioxide. With it, some of the ATP is generated directly in the reactions, which transform glucose.  


4 0
3 years ago
The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.
Naddika [18.5K]

Answer:

12.6.

Explanation:

  • We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

  • It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

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