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dimulka [17.4K]
3 years ago
11

Which phase(s) would experience a decrease in the rate of dissolution with an increase in temperature?

Chemistry
2 answers:
lukranit [14]3 years ago
8 0
D. both gases and liquids because as it disolutes the temp increases

Vitek1552 [10]3 years ago
5 0
I woul say that it would have to be gases
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Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms
Licemer1 [7]
To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
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7 0
3 years ago
a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?
Zina [86]

Answer:

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ =  70 KPa × 600 mL × 313 K / 293K ×150 mL  

P₂ = 13146000 KPa .mL. K /43950 K.mL

P₂ = 299.11 KPa

4 0
2 years ago
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lozanna [386]
Use the ideal gas formula-----> PV= nRT

P= 2.50 atm
V= 250 mL= 0.250 L
n= 0.100 moles
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T= PV/nR

T= (2.50 atm x 0.250 L) / (0.100 moles x 0.0821)= 76.1 K
7 0
3 years ago
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