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3 years ago

The middle layer of earth’s atmosphere is the

Chemistry

2 answers:

netineya [11]3 years ago

6 0

If I remember right, the middle layer is the mesosphere.

Hope this helps! :)

Anni [7]3 years ago

6 0

The middle layer of earth's atmosphere is found in the stratosphere, which is the mesosphere

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Use the periodic table to answer this question. Decomposing calcium carbonate yields calcium oxide and carbon dioxide. What info

erik [133]

This may help you

First write and balance the equation, being: CaCO3 - CaO + CO2 Then, using the periodic table, find the molecular masses of CaCO3 and of CaO, finding their ratio. That will be 100g:56g or 0.1kg:0.056kg. Since you have 4.7kg of CaCO3, it corresponds to Xkg of CaO. Making x the subject, it should be X= 4.7*0.056/100=0,002632

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3 years ago

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Which of the following is a mixture? steel water oxygen gold

Natasha_Volkova [10]

Water is the answer out of all the other options

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3 years ago

How far up can a 200 N elevator be lifted with 600 j of energy

nekit [7.7K]

A 50 kg student runs up a flight of stairs that is 6 m high. How much work is done? 3000 J. Calculate the work done when a force of 1 N moves a book 2 m. 2 J .... 200 W, 100 W. Calculate the power expended when a 500. N barbell is lifted 2.2 m in 2.0 seconds. 550 W. energy. the property of an object or system that ...

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3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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3 years ago

this is the chemical formula for methyl tert-butyl ether (the clean-fuel gasoline additive mtbe):a chemical engineer has determi

Alex17521 [72]

The Number of moles required for MTBE is 2 as per molarity based chemistry theories.

A flammable liquid with a distinct, unpleasant smell is called methyl tert-butyl ether (MTBE).
Since the 1980s, it has been added to unleaded gasoline as a fuel additive to promote more effective burning.
It is created by mixing chemicals like isobutylene and methanol. Gallstones can be removed with MTBE as well.
In this type of treatment, surgically inserted special tubes are used to deliver MTBE directly to the gall bladders of the patients.
Methyl tert-butyl ether is a colorless liquid with a distinct anesthetic-like smell. Vapors are narcotic and heavier than air. 131 °F boiling point. 18 °F is the flash point.
It is miscible in water and less dense than water. Boosts the octane of gasoline.
Therefore it has 2 moles of oxygen.

To study about isobutylene -

brainly.com/question/8409160

#SPJ4

6 0

1 year ago