A(n) _______________ can be formed by linking together several monosaccharides via glycosidic bonds.

Answers

Answer 1

Answer:

Answer:

A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.

Explanation:

Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.

An example of a polysaccharide is starch, or glycogen.

Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.

Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.

Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.

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100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of the oxygen gas is 32 g/mol

$\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}$

Step 3; Convert the temperature to degree Kelvin

$\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}$

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

$\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}$

Hence, the volume of the oxygen gas is 17.5 L

Radioactive gold-198 is used in the diagnosis of liver problems. the half-life of this isotope is 2.7 days. if you begin with a sample of 8.1 mg of the isotope, how much of this sample remains after 2.6 days?

Answers

Answer:

See explanation below

Explanation:

To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:

m = m₀e^-kt (1)

In this case, k will be the constant rate of this element. This is calculated using the following expression:

k = ln2/t₁/₂ (2)

Let's calculate the value of k first:

k = ln2/2.7 = 0.2567 d⁻¹

Now, we can use the expression (1) to calculate the remaining mass:

m = 8.1 * e^(-0.2567 * 2.6)

m = 8.1 * e^(-0.6674)

m = 8.1 * 0.51303

m = 4.16 mg remaining

Final answer:

The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.

Explanation:

This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.

So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.

Learn more about half-life here:

brainly.com/question/31375996

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How to prepare ethanoic acid from ethane

Answers

anonymous
anonymous 5 years ago
First you chlorinate it in presence of light.
C2H6 + Cl2 ---hv -> C2H5Cl + HCl
Then you add aqeuos KOH to get C2H5OH
C2H5Cl+KOH-> C2H5OH+ KCl.
Then you add KMnO4 to get the rquired compound.
C2H5OH ----KMnO4 ---> CH3COOH.

C2h6 + O2 ----> CH3CooH

A 3.25 L solution is prepared by dissolving 285 g of BaBr2 in water. Determine the molarity.

Answers

Answer:

0.295 mol/L

Explanation:

Given data:

Volume of solution = 3.25 L

Mass of BaBr₂ = 285 g

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of solute:

Number of moles = mass/ molar mass

Molar mass of BaBr₂ = 297.1 g/mol

Number of moles = 285 g/ 297.1 g/mol

Number of moles= 0.959 mol

Molarity:

M = 0.959 mol / 3.25 L

M = 0.295 mol/L

Final answer:

The question is about calculating the molarity of a solution. First, convert the given mass of solute into moles using the molar mass. Then, using the molarity formula, divide the moles of solute by the volume of the solution in liters.

Explanation:

In order to determine the molarity of the solution, we will divide the amount of solute (in moles) by the volume of the solution (in liters). The formula for molarity (M) is:

M = moles of solute/volume of solution in liters

First, we need to convert the mass of BaBr2 into moles. The molecular weight of BaBr2 is 297.14 g/mol. So, 285 g of BaBr2 is equal to 285/297.14 = 0.959 moles.

The volume of the solution is given as 3.25 L. So, plugging these values into the formula gives us the molarity of the solution:

M = 0.959 moles/3.25 L = 0.295 M

So, the molarity of the solution is 0.295 M.