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Mamont248 [21]
2 years ago
8

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

Chemistry
1 answer:
leonid [27]2 years ago
6 0
  • Volume=250mL
  • Density=0.71g/ml

\boxed{\sf Density=\dfrac{Mass}{Volume}}

\\ \sf{:}\implies Mass=Density(Volume)

\\ \sf{:}\implies Mass=0.71(250)

\\ \sf{:}\implies Mass=177.5g

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How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Alex787 [66]

Answer:

about 4.8kJ

Explanation:

6 0
2 years ago
Magnesium hydroxide (Mg(OH)2): g/mol
Eva8 [605]

To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.

Mg = 24.305 x 1 = 24.305 amu

O = 15.9994 x 2 =31.9988 amu

H = 1.0079 x 2 = 2.0158 amu

 

Then, add all the components to get the atomic mass of the molecule.

24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu


The atomic mass is just equivalent to its molar mass.

So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.

8 0
3 years ago
Read 2 more answers
Substance A has the following properties.
givi [52]

A curve of temperature vs. time for the entire heating process.

The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.

i ) Calculating the heat and time required to heat the sample to its boiling point:

Boiling point = 85°C

C(liquid) = 2.5 J/g °C

The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,

Therefore, T f = 85°C  and T i = - 20°C

Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,

q3 = 25 g ×  2.5 J/g °C × [85 - (-20)]°C

     = 25 J/°C ×[85+20]°C

     = 6562.5 J

The total heat energy required for heating the sample from initial temperature to boiling point is:-

q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J

                    = 11562.5 J

The Rate of heating = 450 J/min

450. J = 1 min

   11562.5 J = ? min

11562.5 J × 1min/450 J = 25.69 min

ii) Calculating the heat and time required to boil the sample:

∆H Vap = 500 J/g

The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined

q4= m × ∆Hvap

    = 25 g × 500 J/g

   = 12500 J

Thus, total heat required to this phase change is q1 + q2 + q3 + q4  = 500 J + 4500 J +6562.5  J + 12500 J = 24062.5 J

The Rate of heating = 450 J / min

450 J = 1 min

24062.5 J = ? min

24062.5J ×  1min / 450 J = 53.47 min

iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C

C gas = 0.5 J / g °C

The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,

Therefore, T f = 100.°C  and T i = 85°C

q5 = 25 g ×  0.5 J / g °C × [100 - 85] °C

    = 25 J / °C ×15 °C

    = 187.5  J

The total heat energy required for heating the sample from initial temperature to 100°C is

q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J

                                      =24250 J

The Rate of heating = 450 J / min

  450. J = 1 min

 24250 J=? min

Thus, heating the sample to 100.°C takes a total of 53.89 min.

iv) Draw a curve of temperature vs. time for the entire heating process:-

Temperature °C     Temperature K     Heat energy (J)     Time (min)

 -40 °C                       233                             0                     0

-20 °C                          253                          500                  1.11    

Melting -20 °C             253                        5000                   11.11

85 °C                         358                         11562.5              25.69

Boiling 85 °C             358                           24062.5          53.475              

100  °C                       373                             24250          53.89

Hence, the graph for the result is in the image.

Learn more about temperature here:-brainly.com/question/24746268

#SPJ4

4 0
1 year ago
What mass of KNO, will dissolve in 100 g of water at 100°C?
Sphinxa [80]

Answer:

About 170-180 grams of potassium nitrate are completely dissolved in 100 g.

Explanation:

Hello!

In this case, according to the reported solubility data for potassium nitrate at different temperatures on the attached picture, it is possible to bear out that about 170-180 grams of potassium nitrate are completely dissolved in 100 g; considering that the solubility is the maximum amount of a solute that can be dissolved in a solvent, in this case water.

Best regards!

3 0
2 years ago
Wich are the disadvantages of skimming oil spills?
sweet-ann [11.9K]

Oil is sucked up through wide floating heads and pumped into storage tanks. Although suction skimmers are generally very efficient, one disadvantage is that they are vulnerable to becoming clogged by debris and ice and require constant skilled observation.
4 0
3 years ago
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