All categories

3 years ago

What is the room temperature of water?

1 answer:

4 0

I surmise that the person who asked this question was wondering why water in a room may sometimes be colder than ambient air and by how much.

For water to get cooler than ambient air requires two conditions: exposed water surface or water able to seep through a clay pot, and relatively low ambient humidity level (usually lower than 30% relative humidity). These two conditions combined enable evaporation. Now, evaporation of liquid water into water vapor necessitates input of latent energy (2257 kJ/kg), and this is at the expense of the remaining liquid water. Thus, liquid water in a room can be at a temperature lower than room temperature.

<span>By how much? There is no straight answer to this because the temperature of the remaining liquid water depends on (1) the rate of evaporation, which itself depends on the amount of surface exposure and the room humidity and temperature, (2) the amount of water remaining in the pot, and (3) the rate of reheating of the water by the ambient air. In practice, the water temperature can be lower than room temperature by several degrees Celsius. </span>

You might be interested in

Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat

Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

Turnover number = $K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}$

$V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}$

Therefore, moles of $CO_{2}$ hydrolyzed is as follows.

Moles of $CO_{2}$ hydrolyzed = $\frac{Mass of CO_{2}}{Molar mass of CO_{2}}$

= $\frac{0.30}{44}$

= 0.00682 moles

Now, moles of $CO_{2}$ hydolyzed per second is calculated as follows.

Moles of $CO_{2}$ hydolyzed per second = $\frac{0.00682}{60}$

= $1.137 \times 10^{-4} moles/second = V_{max}$

And,

Moles of enzyme = $\frac{Mass}{\text{Molar mass}}$

= $\frac{10.0 \mu g}{30000}$

= $3.33 \times 10^{-10}$ moles

Therefore, the value of $K_{cat}$ is as follows.

$K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}$

= $0.3411 \times 10^{6}$ per second

= $0.3411 \times 60 \times 10^{6}$ per minute

= $20.466 \times 10^{6}$ per minute

Thus, we can conclude that the turnover number ( $K_{cat}$ ) of carbonic anhydrase (in units of $min^{-1}$ ) is $20.466 \times 10^{6}$ per minute.

6 0

3 years ago

Label each statement below as either describing the limit of detection (LOD) or the limit of quantitation (LOQ

Orlov [11]

Answer:

It is the minimum amount of analyte that produces a signal which can be measured with reasonable accuracy - LOQ

The concentration is equal to three times the standard deviation of the signal from the blank divided by the slope of the calibration curve - LOD

The concentration is equal to 10 times the standard deviation of the signal from the blank divided by the slope of the calibration curve - LOQ

It is the minimum amount of analyte that produces a signal that is significantly different from the blank - LOD

Explanation:

We define the limit of detection has the lowest amount of analyte that produces a signal that is significantly different from a blank solution ( the absence of the substance). It is calculated as three times the standard deviation of the signal from the blank divided by the slope of the calibration curve.

The limit of quantitation (LOQ) is defined as the minimum amount of analyte that produces a signal which can be measured with reasonable accuracy. It is measured as 10 times the standard deviation of the signal from the blank divided by the slope of the calibration curve.

6 0

3 years ago

Which best describes what is made of matter

avanturin [10]

Atoms, and everything is made of matter.

5 0

3 years ago

Read 2 more answers

Calculate the Experimental Molar Volume in L/mol of the Hydrogen gas, H2, if the volume of H2 at STP is 52.8 mL and the mass of

lana66690 [7]

Answer:

The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Explanation:

We are given that

Volume of H2 at STP=52.8mL

Mass of magnesium metal ,M(Mg)=0.055g

We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.

Molar mass of Mg=24.305 g/mol

Number of moles= $\frac{given\;mass}{molar\;mass}$

Using the formula

Number of moles of Mg= $\frac{0.055}{24.305}$ moles

Number of moles of Mg=0.00226moles

Number of moles of Mg=Number of moles of H2

Number of moles of H2=0.00226moles

Molar volume of Hydrogen gas (H2)= $\frac{volume\;at\;STP}{No\;of\;moles\;H_2}$

Molar volume of Hydrogen gas (H2)= $\frac{52.8}{0.00226}mL/mol$

Molar volume of Hydrogen gas (H2)= $\frac{52.8}{0.00226}\times 10^{-3}L/mol$

$1L=1000mL$

Molar volume of Hydrogen gas (H2)=23.36L/mol

Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

6 0

2 years ago

What are the isomers of c3h6br2...

maks197457 [2]

C3H6Br2 2 isomers:- bominated allkane

CH3CH2CHBr2 1,1-dibromopropane
<span>CH3CHBrCH2Br 1,2-dibromopropane

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>

7 0

3 years ago