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nirvana33 [79]
3 years ago
6

Glucose (c6h12o6) has a single carbonyl group (-c=o) in its linear form. based on the number of oxygen atoms in glucose, how man

y hydroxyl groups (-oh) would you expect glucose to have?

Chemistry
1 answer:
sveta [45]3 years ago
3 0
Glucose is a hexose monosaccharide. It is one of the three major monosaccharides along with fructose and galactose. These are carbohydrates with a general formula of Cₓ(H₂O)ₓ, where x could be any number.

Now, you don't have to know the structural formula of the glucose to answer this. Just account all the elements in the glucose. You know that there are 6 oxygen atoms all in all. One of them belongs to the single carbonyl group. Consequently, that would mean that the remaining 5 oxygen atoms bond with hydrogen atoms to form 5 OH groups.

Just to be sure let us refer to the structural formula of glucose shown in the picture. It indeed has 5 OH groups.

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a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa
lesya692 [45]

Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

Where:

V₁ = Initial volume

V₂ = Final volume = V₁/2

P₁ = Initial pressure = 832 torr

P₂ = Final pressure  = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂  = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

6 0
3 years ago
Read 2 more answers
How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc
lbvjy [14]

Answer:

249 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

V = 249 L

5 0
3 years ago
Suppose you have a dozen carbon atoms, a dozen gold atoms, and a dozen iron atoms. Even though you have the same number of each,
mixer [17]

Answer:

By weight they have the same mass, but the number of atoms is different

Explanation:

3 0
3 years ago
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How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
bezimeni [28]

Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

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3 years ago
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They are stronger than hydrogen bonding forces.

3 0
3 years ago
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