Answer: The new volume of cake is 1.31 mL.
Explanation:
Given:
= 0.20 mL, 
= ?, 
Formula used to calculate new volume is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that the new volume of cake is 1.31 mL.
The temperature will change from 100K to 173.87 K
calculation
by use of law that is V1/T1=V2/T2
V1=3.75 L
T1=100k
V2=6.53 L
T2=?
make T2 the subject of the formula
T2=(V2 xT1)V1
=6.52 x100/3.75=173.87K
Amines are derivatives of
Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;
Primary Amines: In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.
Secondary Amines: In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.
Tertiary Amines: In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.
Below are three isomers of tertiary amines with molecular formula
C₅H₁₃N.
Answer:
calcium phosphate has the formula Ca3(PO4)2, which has the mass of 310 grams/mol.
1 mol contains 310 grams
2.3*10^-4 moles contain 2.3*10^-4 * 310, which means 713*10^-4 grams, or 71.3 milligrams.
If you wrote the formula right and named the compound wrong, all you have to do is replace 310 with 278 and the answer will be 639.4*10^-4 grams, or 63.94 milligrams.
<span>n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9
First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4.
We know that 1dm3=1L, so H2SO4's molarity is
C=nV=18.0moles1.0L=18M
In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so
18.0moles1Lâ‹…98.0g1mole=1764g1L
Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution
98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolution→
masssolution=1764gâ‹…100.0g98g=1800g
Therefore, 1L of 98wt% H2SO4 solution will have a density of
Ď=mV=1800g1.0â‹…103mL=1.8gmL
H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be
cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m
Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that
100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4
100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O
So, H2SO4's mole fraction is
molefractionH2SO4=11+0.11=0.9</span>