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3 years ago

A sugar that has the chemical formula C5H10O5 would be characterized as _____.

Chemistry

1 answer:

Tanya [424]3 years ago

7 0

Answer:

The answer to your question is: a. pentose

Explanation:

a) a pentose is a monosaccharide that has 5 carbons in its structure, if we look at the question the chemical formula only has 5 carbons, then this is the right answer.

b) an oligosaccharide is a group of 2 or more monosaccharides (till 5 or 6) then, it will have 10 or more carbons, this is not the right answer

c) a triose is a monosaccharide that has only three carbons, this is not the right answer.

d) a hexose is a monosaccharide that has 6 carbons, of course this answer is wrong.

e) a polysaccharide is a group of 6 or more monosaccharides, it will have 20 or more carbons, this answer is wrong.

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The hydrolysis of sucrose (C12H22O11) into glucose and fructose in acidic water has a rate constant of 1.8×10−4s−1 at 25 ∘C. Ass

Musya8 [376]

Answer:

Mass of sucrose that is hydrolyzed = 123.55 g

Explanation:

Since the reaction is a first ofree reaction,

Let the initial concentration of sucrose be C₀

And the concentration of sucrose left at any time be C.

r = (dC/dt) = -kC (Minus sign because it's a rate of reduction)

K = rate constant

(dC/dt) = - kC

(dC/C) = -kdt

∫ (dC/C) = -k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from C₀ to C and the Right hand side from 0 to t.

We obtain

In (C/C₀) = -kt

(C/C₀) = (e⁻ᵏᵗ)

C = C₀ e⁻ᵏᵗ

So, we can find the concentration of sucrose left at the given time

k = 1.8 × 10⁻⁴ s⁻¹

t = 200 minutes = 200 × 60 = 12000 s

kt = 12000 × 1.8 × 10⁻⁴ = 2.16

C₀ = 0.160 M

C = 0.160 e⁻²•¹⁶ = 0.01845 M

So, the concentration that has been used up = 0.160 - 0.01845 = 0.142 M

To calculate the mass of sucrose that has reacted, we'll convert the concentration used up into number of moles.

Concentration = (number of moles)/(volume in Litres)

number of moles = concentration × volume in Litres

Volume in Litres = 2.55L

Number of moles = 0.142 × 2.55 = 0.361 moles

Then, mass = number of moles × Molar mass

Molar mass of sucrose = 342.3 g/mol

Mass of sucrose used up = 0.361 × 342.3 = 123.55 g

6 0

2 years ago

6. Suppose you are going to measure the length of a pencil in centimeters.

Montano1993 [528]

Answer:

Their answers must be in two decimal places

3 0

2 years ago

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

Energy that is lost as heat

Tema [17]

Can still be used to do work

3 0

2 years ago

WILL MARK BRAINLIEST !! HELP ASAP

andre [41]

I am almost positive that the answer is D

6 0

3 years ago

Read 2 more answers