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3 years ago

¿que diferencia hay entre psicrómetro e higrómetro?

Chemistry

1 answer:

Tasya [4]3 years ago

7 0

Answer:

Un higrómetro que para calcular la humedad se vale de la diferencia de temperaturas entre un termómetro con el bulbo seco y otro con el bulbo húmedo, normalmente se denomina psicrómetro.

Explanation:

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The numbers of spiders in the house and the outside temperature have an indirect relationship.what will happen to the spiders in

aalyn [17]

Indirect relationship: relationship between two variables which move in opposite directions.

If temperature decreases, number of spiders would increase.

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3 years ago

A gas is held under conditions of standard temperature and pressure. It is found that 44.0 grams of the gas occupies a volume of

storchak [24]

Answer:

CO2

Explanation:

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At STP 1 mol=22.4 Liters, so we now know that it is asking for which of the gasses has a molar mass of 44, and CO2 is th only one with that molar mass

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3 years ago

A titration is completed using 10.0 mL of 0.50 M HCl and an unknown concentration of NaOH solution. If the equivalence point is

Drupady [299]

Answer:

0.172 M

Explanation:

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3 years ago

How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with

Kisachek [45]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = $K_a=6.8\times 10^{-4}$

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaF]}{[HF]}$

Now put all the given values in this expression, we get:

$2.60=3.17+\log (\frac{[NaF]}{0.0310})$

$[NaF]=0.00834M$

Now we have to calculate the moles of NaF.

$\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole$

Now we have to calculate the mass of NaF.

$\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g$

Therefore, the mass of sodium fluoride added should be 0.105 grams.

4 0

3 years ago

A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at

lawyer [7]

<u>Answer:</u> The final temperature of the system is 14.60°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 25.00 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of aluminium = 100°C

$T_2$ = initial temperature of water = 10°C

$c_1$ = specific heat of aluminium = 0.900 J/g°C

$c_2$ = specific heat of water= 4.18 J/g°C

Putting values in equation 1, we get:

$25\times 0.900\times (T_{final}-100)=-[100\times 4.18\times (T_{final}-10)]$

$T_{final}=14.60^oC$

Hence, the final temperature of the system is 14.60°C

3 0

3 years ago