Answer:
P₂ = 1312.88 atm
Explanation:
Given data:
Initial temperature = 25°C
Initial pressure = 1250 atm
Final temperature = 40°C
Final pressure = ?
Solution:
Initial temperature = 25°C (25+273.15 = 298.15 K)
Final temperature = 40°C ( 40+273.15 = 313.15 k)
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
1250 atm / 298.15 K = P₂/313.15 K
P₂ = 1250 atm × 313.15 K / 298.15 K
P₂ = 391437.5 atm. K /298.15 K
P₂ = 1312.88 atm
Answer:
11.31 g.
Explanation:
Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L)).
<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>
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<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>
Answer:
oh coooool
I have one Id it is on ace with maybe 5800 points
maybe this picture will help in something
Separation will be achieved if one component adheres to the stationary phase more than the other component does.
<em>Acetic acid, HC2H3O2</em>
First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g
Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
P1 = ((4)(1)/60)(100%) = <em>6.67%</em>
<em> Carbon:</em>
P2 = ((2)(12)/60)(100%) = <em>40%</em>
<em>Oxygen</em>
P3 =((2)(16) / 60)(100%) = <em>53.33%</em>
<em>Glucose, C6H12O6</em>
The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180
The percentages of the elements are as follow,
<em> Hydrogen:</em>
P1 = (12/180)(100%) = <em>6.67%</em>
<em>Carbon:</em>
P2 = ((6)(12) / 180)(100%) = <em>40%</em>
<em>Oxygen:</em>
P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>
b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.