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3 years ago

Distinguish between sugar and non-sugar with examples.

Chemistry

1 answer:

kap26 [50]3 years ago

3 0

Answer:

Reducing sugars are sugars where the anomeric carbon has an OH group attached that can reduce other compounds. Non-reducing sugars do not have an OH group attached to the anomeric carbon so they cannot reduce other compounds. ... Maltose and lactose are reducing sugars, while sucrose is a non-reducing sugar

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013 10.0 points

34kurt

Answer:

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Explanation:

5 0

2 years ago

Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

The Brahman cattle have good resistance to high temperature, but its meat is poor and tough to eat. The English shorthorn cattle

N76 [4]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurate

Over [174]

Answer:

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25)/ 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg

4 0

2 years ago

How many grams of CH4 will be in a 500ml contain at STP?

ArbitrLikvidat [17]

Answer:

Mass = 0.32 g

Explanation:

Given data:

Mass of CH₄ = ?

Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)

Temperature = 273 K

Pressure = 1 atm

Solution:

Volume of CH₄:

500 mL (500 mL× 1L/1000 mL= 0.5 L)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

By putting values,

1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K

0.5 atm.L = n×22.4 atm.L/ mol

n = 0.5 atm.L / 22.4 atm.L/ mol

n = 0.02 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.02 mol × 16 g/mol

Mass = 0.32 g

7 0

3 years ago