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kvasek [131]
3 years ago
12

Freezing point of carbon dioxide a.meter b.liter c.kilogram d.kelvin

Chemistry
1 answer:
Orlov [11]3 years ago
7 0

The Freezing point is ALWAYS the same as the melting point,even for different elements. 6,422 is its melting,and freezing point. Same with gold,iron,hydrogen,and many others

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A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
Read 2 more answers
Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction?
Alex73 [517]
Your answer is B.
Because it says that that carbon burns in presence of oxygen (C+O) which is equal ( => ) to Carbon Dioxide (CO_{2})
3 0
3 years ago
Read 2 more answers
How many atoms are in a sample of chromium with a mass of 31 grams? O a 2.4 x 1024 atoms of chromium b 3.6 x 1023 atoms of chrom
lukranit [14]

Answer:

b. 3.66x10²³ atoms of chromium.

Explanation:

First we calculate how many moles are there in 31 grams of chromium, using its molar mass:

  • Molar Mass of Chromium = 51 g/mol (This can be found on any periodic table)
  • 31 g ÷ 51 g/mol =  0.608 mol

Then we <u>calculate how many atoms are there in 0.608 moles</u>, using <em>Avogadro's number</em>:

  • 0.608 mol * 6.023x10²³ atoms/mol = 3.66x10²³ atoms

The correct answer is thus option b. 3.66x10²³ atoms of chromium.

4 0
3 years ago
Noble gas configuration
saw5 [17]

Answer:

Rubidium= [Kr] 5s^1

Calcium= [Ar] 4s^2

Aluminium= [Ne] 3s^2 3p^1

Explanation:

A noble gas configuration begins with the elemental symbol of the last noble gas prior to the atom. The symbol is then followed by the remaining electrons.

Hope this helped! good luck :)

3 0
3 years ago
a sample of an unknown gas has a mass of 0.582g. it's volume is 21.3 mL a a temp of 100 degrees Celsius and a pressure of 754 mm
Zielflug [23.3K]

Answer:

Molar mass = 0.09 × 10⁴ g/mol

Explanation:

Given data:

Mass = 0.582 g

Volume = 21.3 mL

Temperature = 100°C

Pressure = 754 mmHg

Molar mass = ?

Solution:

(21.3 /1000 = 0.0213 L)

(100+273= 373 K)

(754/760 = 0.99 atm)

PV = nRT

n = PV/RT

n = 0.99 atm × 0.0213 L / 0.0821 atm. L. mol⁻¹. k⁻¹ × 373 K

n =0.02 mol/ 30.6

n = 6.5 × 10⁻⁴ mol

Molar mass = Mass/ number of moles

Molar mass = 0.582 g / 6.5 × 10⁻⁴ mol

Molar mass = 0.09 × 10⁴ g/mol

6 0
3 years ago
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