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Pie
4 years ago
12

what is the relative humidity when the air temperature is 28 degrees celcius and the wet bulb temperature is 20 degrees celcius?

Chemistry
2 answers:
Sedaia [141]4 years ago
6 0

Answer:

47 %  

Step-by-step explanation:

The formula for <em>relative humidity</em> (RH) consists of three parts:

a_{d} = 6.112e^{\frac{17.502T_{d} }{240.97+T_{d}}}

a_{w} = 6.112e^{\frac{17.502T_{w} }{240.97+T_{w}}}

RH = \frac{ a_{w}-0.668 745(1+0.00115T_{w})(T_{d}-T_{w})}{a_{d}}\times100

<em>Data: </em>

T_{d} = 28 ^{\circ}\text{C}; T_{w} = 20^{\circ}\text{C}

Calculations:

a_{d} = 6.112e^{\frac{17.502\times28}{240.97+ 28}}

a_{d} = 6.112e^{\frac{490.1}{238.97}}

a_{d} = 6.112e^{1.822}

a_{d} = 6.112\times 6.184

a_{d} = 37.88

a_{w} = 6.112e^{\frac{17.502\times20}{240.97+ 20}}

a_{w} = 6.112e^{\frac{350.0}{260.97}}

a_{w} = 6.112e^{1.341}

a_{w} = 6.112\times 3.824

a _{w} = 23.37

RH = \frac{ 23.37 - 0.668 745(1+0.00115\times20)(28 - 20)}{37.88}\times100

RH = \frac{ 23.37 - 0.668 745(1+0.02300)\times8}{37.88}\times100

RH = \frac{ 23.37 - 0.668 745\times1.02300\times8}{37.88}\times100

RH = \frac{ 23.37 - 5.4738}{37.88}\times100

RH = \frac{ 17.90}{37.88}\times100

RH = 47.3 \%

It's much easier to use a <em>RH table</em> (see below) and get the same result.

Tanya [424]4 years ago
4 0

10 i think not sure

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Can someone help me and explain why they got what they got?
allsm [11]

Answer:

24.32 amu

Explanation:

From the question given above, the following data were obtained:

Isotope A (Mg–24):

Mass of A = 24 amu

Abundance (A%) = 79%

Isotope B (Mg–25):

Mass of B = 25 amu

Abundance (B%) = 10%

Isotope C (Mg–26):

Mass of C = 26 amu

Abundance (C%) = 11%

Average atomic mass of Mg =?

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

= [(24 × 79)/100] + [(25 × 10)/100] + [(26 × 11)/100]

= 18.96 + 2.5 + 2.86

= 24.32 amu

Thus, the average atomic mass of Mg is 24.32 amu

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3 years ago
Which statement is false?
emmasim [6.3K]

Answer:

  • <u><em>The first statement is false:  a.At equilibrium, equal amounts of products and reactants are present. ΔG° is a function of Keq.</em></u>

Explanation:

When one part of a statement is false, the whole statement is false.

At <em>equilibrium,</em> the amounts of products and reactants does not have to be equal.

At equlibrium the rates of the forward reaction and the reverse reaction must be equal.

An equilibrium reaction may be represented by:

  • A + B ⇄ C + D

That represents two reactions:

  • Direct reaction: A + B → C + D (A and B yield C and D)

  • Reverse reaction: A + B ← C + D (C and D yield A and B: note that the arrow goes from right to left)

So, it is when the direct and the forward rates are equal that there is not net change in the amounts of all the species and so the reaction is is equilibrium).

As per the other statement, both parts are true:

  • When reactants become products, they do so through an intermediate transitrion state: when the reactants approach each other and collide with enough energy and appropiate position, the bonds start to break and the bonds of the products start to form. This is the transition state.

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3 years ago
How many sulfur atoms are in H2SO4?
NemiM [27]

Answer: 1

Explanation:

6 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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In which way are a prokaryotic cell and a eukaryotic cell alike?  
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Prokaryotic cells do not contain membrane-closed organelles, (??) and they are not the same size at all. (prokaryotic are way smaller) Just to help narrow down, if I'm right! (: Sorry, I know that's not completely helpful.
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3 years ago
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