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3 years ago

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7. A phase change occurs when the _ energy of the particles is high enough to overcome the ___ forces between th

e particles.

1 answer:

love history [14]3 years ago

6 0

I at least know the first blank is kinetic

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Which of the following could result from having few or no ?-1,6 linkages in glycogen? Select all that apply.

Katen [24]

Answer:

The correct answer is 1 glycogen degradation would slow down.

Explanation:

Glycogen is the principle storage polysaccharide present in the liver and muscle of human body.

Glycogen contain both alpha-1,4-glycosidic linkage and alpha -1,6-glycosidic linkage.During glycogenolysis some glucose residues are transferred from branch point of the glycogen to its end and thereafter a single glucose residue is linked to the branch point of glycogen by alpha-1,6-glycosidic linkage.

The alpha-1,6-glycosidic linked glucose of glycogen is finally get separated from glycogen by the catalytic activity of alpha-1,6-glycosidase enzyme in the final step of glycogenolysis.

According to the given question if there is no alpha-1,6-glycosidic linkage in the glycogen then glycogen degradation will slow down.

5 0

3 years ago

2HCl+Ca(OH)2⟶CaCl2+2H2O

RUDIKE [14]

Isn’t it just 3 mols?

5 0

3 years ago

1. What is the Charlotte?<br> A. A train<br> B. A car<br> D. An airplane<br> C. A ship

dybincka [34]

Answer:

B is the Charlotte

Explanation:

it is Charlotte because it has 4 wheels

4 0

3 years ago

The number of electrons in n=1 and n=2 shells of aluminum are

harina [27]

Answer:

n=1 holds two electrons and n=2 holds eight electrons.

Explanation:

Hello

In this case, since the atomic number of aluminum is 13, its electron configuration is:

$Al^{13}: 1s^2,2s^2,2p^6,3s^2,3p^1$

In such a way, we can see that the level n=1 is filled with two electrons since the subshell s is able to hold two electrons and the level n=2 is also filled but with eight electrons as s holds two whereas p holds 6. Moreover, n=3 is holding three electrons.

Best regards.

5 0

3 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago