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3 years ago

Which statements are correct relative to carbohydrates?

Chemistry

1 answer:

inna [77]3 years ago

8 0

Answer:

Hi, the statements given in the question (a and b) are combination of right and wrong answers. But i will split the statement a and b into necessary fragments.

Carbohydrates are quick source of energy relatively. That is the only correct statement in the question.

Explanation:

Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen with the general formula (Cx H2y Oy) or Cx (H2O)y. They can be classified into simple sugar or complex sugar.

Simple sugars can be further divided into Monosaccharides (e.g glucose, fructose and galactose) and Disaccharides (e.g sucrose and maltose).

Complex sugar can be referred to as Polysaccharides (e.g cellulose and starch)

The statement above that says carbohydrate contains carbon, hydrogen and nitrogen polymers is not correct.

Carbohydrates are not mostly monosaccharides. They also contain Disaccharides and Polysaccharides.

They are not the only source of fuel.

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Who was the 33rd president of the United States?

Orlov [11]

Answer:

Harry S. Truman

Explanation:

Because he was the 33rd president

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3 years ago

Carbon-14 (14C) dating assumes that the carbon dioxide on the Earth today has the same radioactive content as it did centuries a

Nataly [62]

Answer: The tree was burned 16846.4 years ago to make the ancient charcoal

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5715 years

Putting values in above equation, we get:

$k=\frac{0.693}{5715yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = 13 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{13}\\\\t=16864.4yrs$

Hence, the tree was burned 16846.4 years ago to make the ancient charcoal

8 0

2 years ago

What property is used to calculate the ph of a solution

icang [17]

A. The concentration is in mol/L

7 0

3 years ago

Convert the following Grams: 0.200 moles of H2S

ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

0.200 mol H₂S

Step 2: Identify Conversions

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

Step 3: Convert

Set up: $\displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})$
Multiply: $\displaystyle 6.818 \ g \ H_2S$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

6.818 g H₂S ≈ 6.82 g H₂S

7 0

3 years ago

What is the PH of a 1.3×(10)^-9 M HBr solution?

olga nikolaevna [1]

pH solution = 8.89

<h3>Further explanation</h3>

Given

The concentration of HBr solution = 1.3 x 10⁻⁹ M

Required

the pH

Solution

HBr = strong acid

General formula for strong acid :

[H⁺]= a . M

a = amount of H⁺

M = molarity of solution

HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1

Input the value :

[H⁺] = 1 x 1.3 x 10⁻⁹

[H⁺] = 1.3 x 10⁻⁹

pH = - log [H⁺]

pH = 9 - log 1.3

pH = 8.89

7 0

2 years ago