Here, we are going to calculate the mass % of C in the mixture.
What is a Mixture?
A mixture is composed of one or more pure substances in varying composition. There are two types of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, while homogeneous mixtures appear uniform throughout.
Given that,
The mass % of CO =35.0% =35.0 g in 100 g mixture
The mass % of CO2 = 65% =65 g in 100 g mixture
Therefore,
The mass of C from CO = 15.007 g C
Similarly,
The mass of C from CO2 = 17.738 g C
Thus, the total mass of C = 15.007 g+17.738 g =32.745 g
Therefore,
The mass % of C= 32.745% =32.7%
Thus, the mass % of C in the mixture is 32.7%
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The initial temperature is 137.34 °C.
<u>Explanation:</u>
As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.
Q = mcΔT
So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.
ΔT =
ΔT =
Since the difference in temperature is 2.66, then the initial temperature will be
Final temperature - Initial temperature = Change in temperature
140-Initial temperature = 2.66
Initial temperature = 140-2.66 = 137.34 °C
Thus, the initial temperature is 137.34 °C.
Answer:
1) C2H4(OH)2
Explanation:
A 1,2-ethanediol has an ethane structure consisting of two Carbon atoms with a hydrogen from each carbon substituted by a hydroxyl group. This makes it a 1,2-diol.
Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Answer:
46
Explanation:
Sodium metal has a molar mass of
22.99
