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Lelu [443]
3 years ago
7

Which of the following statements regarding glucose is FALSE?a) Glucose is the main component of starch and glycogen.b) Glucose

is also called blood sugar.c) Glucose is the most common disaccharide in our diet.d) Glucose is the most important carbohydrate fuel for the human body.
Chemistry
1 answer:
marishachu [46]3 years ago
8 0

Answer:

The correct option is C.

Explanation:

Carbohydrates are one of the macro molecules that are consumed by living organisms. The end product of carbohydrate is glucose. Glucose is a very important fuel that the body cells used to produce energy, which they use to carry out their daily activities. Glucose is also known as blood sugar and it is the only fuel that living cells can use for the production of ATP. Other food macro molecules such as lipids and proteins can also be converted to glucose if there is a need for that. Glucose is always stored in the body in form of glycogen.

The statement given in option C about glucose is wrong because glucose is a monosaccharide and not a disaccharide.

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At a certain temperature this reaction follows second-order kinetics with a rate constant of Suppose a vessel contains at a conc
Norma-Jean [14]

The question is incomplete, here is the complete question:

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹

2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.

<u>Answer:</u> The concentration of SO_3 in the vessel after 0.240 seconds is 0.24 M

<u>Explanation:</u>

For the given chemical equation:

2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 14.1M^{-1}s^{-1}

t = time taken= 0.240 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 1.44 M

Putting values in above equation, we get:

14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)

[A]=0.245M

Hence, the concentration of SO_3 in the vessel after 0.240 seconds is 0.24 M

4 0
3 years ago
What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu
Arisa [49]

Answer:

\boxed {\boxed {\sf P_2= 9 \ atm}}

Explanation:

Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

P_1V_1=P_2V_2

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

3.0 \ atm * 150 \ mL = P_2V_2

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

3.0 \ atm *150 \ mL= P_2 * 50 \ mL

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

450 \ atm*mL= P_2 * 50 \ mL

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.

\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}

\frac{450 \ atm*mL}{50 \ mL}= P_2

The units of milliliters will cancel.

\frac{450 \ atm}{50 }= P_2

9 \ atm =P_2

The new pressure is <u>9 atmospheres.</u>

8 0
3 years ago
PLZZ HELLLPBalance the following reaction: AL2O3 + HCI ⇒ ALCI3 + H2O
Alina [70]
Im assuming that the 2 is a sub of the Al and not the # for O  and that u made a type of AL which is Al
its been a while but 

Al2O3 + 6HCl  = 2ALCl3 + 3H2O

3 0
3 years ago
A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so
uranmaximum [27]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>

Explanation:

[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>

Ionic equation:

{ \bf{2OH { }^{ - }  _{(aq)} + 2H { }^{ + } _{(aq)}→  2H _{2} O _{(l)} }}

first, we get moles of potassium hydroxide in 28.94 ml :

{ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

{ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\  = { \underline{423.95 \: g \approx424 \: grams}}

for the molar mass:

8 0
2 years ago
For many purposes we can treat methane (CH) as an ideal gas at temperatures above its boiling point of - 161. °C. Suppose the te
AysviL [449]

Answer:

Explanation:

T₁ = 100 + 273 = 373K

T₂ = 273 + 119 = 392 K

V₁ = initial volume

V₂ = Final volume

P₁ = P₁

P₂ = .85P₁

Using gas law equation

\frac{P_1\times V_1}{T_1} =\frac{P_2\times V_2}{T_2}

= \frac{P_1\times V_1}{373} =\frac{.85P_1\times V_2}{392}

V₂ = 1.236 V₁

% increase in volume

= V₂-V₁ / V₁  x 100

= (1.236 V₁ - V₁ / V₁)x 100

= .236 x 100

= 23.6 % .

5 0
3 years ago
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