What factors affect the rate of dissolving a solid in a liquid? Explain...

3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on

AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.

3 0

3 years ago

Which of the following is NOT a true statement?

Artyom0805 [142]

Answer:

c is not a true statement

8 0

3 years ago

Read 2 more answers

How do molecular bonds form?

vaieri [72.5K]

Answer:

"A molecular, or covalent bond, is formed when atoms bond by sharing pairs of electrons. This sharing can occur from atom to atom, or from an atom to another molecular bond."

Explanation:

Google

8 0

3 years ago

Read 2 more answers

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

2. ATO kg bouting ball would require what force to accelerate down an alleyway at a rate of ons?

Mars2501 [29]

Answer:

2. 30N

3. 5,000N

4. 15 kg

5. 2,800 kg

Explanation:

2. A 10 kg bowling ball would require what force to accelerate down an alleyway at a rate of 3m/s² ?

Notice that I completed the question with the garbled and missing values:

Data:

F = ?

m = 10 kg

a = 3m/s²

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=10kg\times 3m/s^2=30N$

3. Salty has a car that accelerates at 5 m/s². If the car has a mass of 1000 kg, how much force does the car produce?

Notice that I arranged the typos.

Data:

F = ?

m = ?

a = ?

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=1,000kg\times5m/s^2=5,000N$

4. What is the mass of a falling rock if it produces a force of 147 N?

Data:

F = 147N

m = ?

a = falling rock

Physical principles:

neglecting air resistance ⇒ a = g: gravitational acceleration: 9.8m/s²

Newton's second law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 147 N and a = g = 9.8m/s², and compute

$m=\dfrac{147N}{9.8m/s^2}=15Kg$

5. What is the mass of a truck if it produces a force of 14,000 N while accelerating at a rate of 5 m/s²?

Data:

F= 14,000N

m = ?

a = 5m/s²

Physical principles:

Second Newton's law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 14,000 N and a = 5m/s², and compute

$m=\dfrac{14,000N}{5m/s^2}=2,800kg$

6 0

3 years ago