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Aleksandr [31]
4 years ago
10

According to the graph, which statement is true?

Chemistry
1 answer:
damaskus [11]4 years ago
5 0
Nitrogen is the one
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As ________________ increases, air pressure decreases. A) pollution B) water vapor C) radiation D) altitude
lara [203]

Answer:

d or b

Explanation:

8 0
3 years ago
Choose the statement that best describes the effects of a higher or lower pH on enzymes.
matrenka [14]
<h3>Answer:</h3>

Higher or lower pH affect the structure of the enzyme and reduces the enzyme activity.

<h3>Explanation:</h3>
  • Enzymes are biological molecules that speed up the rate of chemical reactions.
  • Enzyme activity is affected by several factors which include temperature, pH, and enzyme inhibitors.
  • Enzymes work on specific molecules known as substrates. They are substrate-specific.
  • Enzymes work best at optimum pH and temperature.
  • Higher or lower pH affects the structure and charges of an enzyme thus affecting enzyme activity.
  • Lower temperature deactivates the enzymes while higher temperatures denature the enzymes.
8 0
3 years ago
Suppose a layer of oil is on the top of a beaker of water. Water in many oils is slightly soluble, so its concentration is so lo
rodikova [14]

Explanation:

It is given that energy to transfer one water molecule is 2.208 \times 10^{-20} J/molecule

As it is known that in 1 mole there are 6.022 \times 10^{23} atoms.

So, energy in 1 mole = 2.208 \times 10^{-20} \times 6.022 \times 10^{23} J/mol

                                  = 13.3 kJ/mol

As,    log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}

Putting the given values in the above formula as follows.

                log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}

               log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}

                  log \frac{k_{2}}{k_{1}} = 0.08377

                       \frac{k_{2}}{k_{1}} = 1.213 = \frac{Concentration_{2}}{Concentration_{1}}

                 Concentration_{2} = 1.213 \times Concentration_{1}

                                             = 1.213 \times 9 \times 10^{-4}

                                             = 10.915 \times 10^{-4} water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is 10.915 \times 10^{-4}.

             

4 0
3 years ago
How many moles of Calcium are in a sample of 40g?
NeTakaya
<span>40 grams ÷ 40.08 grams/moles = 1 mole</span>
5 0
3 years ago
When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/m
andreev551 [17]

Answer:

The heat of vaporization in kJ/mol of methanol

= 3.48 kJ/mole

Explanation:

Here,the heat of vaporization in kJ/mol is asked to calculate.<em><u>This means it is asked to calculate heat required for 1 mole of methanol.</u></em> Since the substance<em> vaporizes </em>, so this is called <u>heat of vaporization.</u>

Divide the thermal energy with the number of moles .

This is simple mathematics :

If 2.50 mol of liquid require = 8.70 kJ of energy

Then, 1 mole will need =

\frac{8.70}{2.50}

= 3.48 kJ/mole

<em><u>Follow the units :</u></em>

The heat of vaporization in kJ/mol of methanol is asked.

Heat\ of\ Vaporization=\frac{Thermal\ energy}{Moles}

=\frac{kJ}{mole}

Unit = kJ /Mole

3 0
3 years ago
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