Answer:
84.259 kPa
Explanation:
We know that:
1 kPa is approximately equal to 7.5 mmHg
Therefore, to convert 632 mmHg to kPa, we will simply use cross multiplication as follows:
1 kPa .................> 7.5006 mmHg
?? kPa ................> 632 mmHg
632 mmHg = (632*1) / (7.5006) = 84.259 kPa
Hope this helps :)
Answer:
The mixture contains 8.23 g of Ar
Explanation:
Let's solve this with the Ideal Gases Law
Total pressure of a mixture = (Total moles . R . T) / V
We convert T° from °C to K → 85°C + 273 = 358K
3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L
(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles
0.756= Total moles from the mixture
Moles of Ar + Moles of H₂ = 0.756 moles
Moles of Ar + 1.10 g / 2g/mol = 0.756 moles
Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206
We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g
Answer:
C, 42g
Explanation:
In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).
Assuming no heat is lost to surroundings,
the energy lost from metal pellet = energy gained for water
Since E = mc∆T
(energy = mass x specific heat capacity x temperature change)
mc∆T (metal pellet) = mc∆T (water)
100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)
3958.96 = 94.14m
m = 42g
Answer:
0.99 kg O₂
1.9 kg SO₂
Explanation:
Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.
S + O₂ → SO₂
The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:
1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂
The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:
1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂
Answer:
D. Freezing
Explanation:
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