Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Science does not have anything to say about right and wrong the meaning of life or the extensive or the subjective
% composition of ethanol = 34.51%
% composition of water = 65.49%
<h3>What is density?</h3>
A material's density is defined as its mass per unit volume.
Given data:
The density of ethanol = 0.7890 g/mL
The density of water = 0.9982 g/mL
The density of mixture = 0.926 g/mL
Let the % composition of ethanol = x
Let the % composition of water = 100-x
Now density of the mixture
%
Hence,
% composition of ethanol = 34.51%
% composition of water = 65.49%
Learn more about the density here:
brainly.com/question/952755
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Answer:
A. is alloyed with carbon and chromium to make stainless steel.
Explanation:
Steel is an alloy formed mainly of iron and carbon but some other metals like chromium are also added in little amounts.
Is steel, the percentage of iron and carbon together is about 90% and the rest metals fall in the 10% part.
Although the cost of steel is low, it has a very high tensile strength and that's why it is used in tools, ships, buildings, trains and in various types of infrastructures.
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
