HELP

Una pieza de silicio tiene un volumen de 8 cm3. ¿Cuál es su masa? *

Maurinko [17]

Answer:

el volumen es igual a masa decidida por densidad

8 0

2 years ago

For test 6 (150 hevy 50 light particles), what is the pressure contribution from the heavy particles?

lawyer [7]

Answer:

The pressure contribution from the heavy particles is 17.5 atm

Explanation:

According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.

In the simulation:

the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,

the total pressure of the mixture of 150 heavy and 50 light particles was measured to be 23.4 atm

Total pressure = partial pressure of Heavy particles + partial pressure of light particles

23.4 atm = partial pressure of Heavy particles + 5.9 atm

Partial pressure of Heavy particles = (23.4 - 5.9) atm

Partial pressure of Heavy particles = 17.5 atm

Therefore, the pressure contribution from the heavy particles is 17.5 atm

4 0

3 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

stira [4]

Explanation:

The given data is as follows.

Mass of apple sauce mixture = 454 kg

Heat added (Q) = 121300 kJ

Heat capacity ( $C_{p}$ ) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$

So, Heat given by heat exchanger = heat taken by apple sauce

Q = $mC_{p} \Delta T$

or, Q = $mC_{p} (T_{f} - T_{i})$

Putting the given values into the above formula as follows.

Q = $mC_{p} (T_{f} - T_{i})$

121300 kJ = $454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)$

$T_{f}$ = $76.5^{o}C$

Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$ .

3 0

3 years ago

Why do teapots have highly polished surface

Natali [406]

Good conductors of electricity and heat, form cations by loss of electrons,

6 0

3 years ago

Read 2 more answers

If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

So

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

3 years ago