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Gnesinka [82]
3 years ago
15

Draw each of the following. Note: If you do not remember the stereocenter configurations for a particular sugar, guess; substant

ial partial credit will be given if the remainder of the structure is correct. (22 points):
a. A dash-wedge structure of any kind of ketopentose.
b. A Fischer projection of L-allose.
c. A Eischer protection of art-mannopyranose.
d. A Haworth projection of p-o-talopyranose.
e. A furanose written in any manner you chose.
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer: its a for AMONG US

Explanation:

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How are the scientific method and the engineering method similar?
aliya0001 [1]
The answer is B

Explanation: Not D because only engineering does that
Plz mark brainliest

3 0
3 years ago
Enter the net ionic equation for the reaction of aqueous sodium chloride with aqueous silver nitrate.
Ksivusya [100]

An ionic equation refers to a chemical equation where the electrolytes in aqueous solution are demonstrated as dissociated ions. Generally, this is a salt dissolved in water, where the ionic species are succeeded by (aq) in the equation, to suggest that they are in aqueous solution.  

Net ionic equation refers to a chemical equation for a reaction that lists only those species that takes part in the reaction. It is an equation, which demonstrates only the reactants taking part in the formation of a precipitate.  

In the given case, balanced chemical equations is,  

AgNO₃ (aq) + NaCl (aq) = AgCl (s) + NaNO₃ (aq)

Complete ionic equation:  

Ag⁺ (aq) +NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) = AgCl (s) + Na⁺ (aq) + NO₃⁻ (aq)

Net Ionic equation:  

Ag⁺ (aq) + Cl⁻ (aq) = AgCl (s) (Silver chloride settles as white precipitate)


6 0
3 years ago
Read 2 more answers
Potassium (K) and calcium fluoride (CaF2) can both be classified as<br> (Please help)
nasty-shy [4]
Potassium and calcium fluoride are both metals
8 0
3 years ago
When 2.50 g of an unknown weak acid (ha) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of
baherus [9]
When dT = Kf * molality * i
                = Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
                               = 2.5g /250g x 1 mol /85 g x1000g/kg
                               =0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution 
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and  i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
     = (0.1176 * 0.359)^2 / (1-0.359)
     = 2.8x10^-3



5 0
3 years ago
what is the  concentration of a NaCl solution, in Molarity, if you add 59.76 g of NaCl into 270 mL of H2O
enot [183]

Hey there!

Molar mass NaCl = 58.44 g/mol

Number of moles

n =  mass of solute / molar mass

n = 59.76 / 58.44

n = 1.0225 moles of NaCl

Volume in liters:

270 mL / 1000 => 0.27 L

Therefore:

M = number of moles / volume ( L )

M = 1.0225 / 0.27

= 3.78 M

Hope that helps!

7 0
3 years ago
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