Shaun’s doctor has recommended that he consume 200 calories of protein a day. Shaun consumed 176 calories of protein in one day. How much more protein does he need to consume to reach the required amount?

Answers

Answer 1
Answer:

Answer:

24 calories of protein

Explanation:

200 - 176 = 24 calories


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what are two conditions atoms must have in order to form covalent bonds with one another

Answers

Covalent bonds are formed mainly between two non metals as electrostatic or ionic bond is prevented from being formed take the example of methane or CH4 (4 means four atoms of hydrogen) in this covalent bonding one atom of carbon combines with four atoms of hydrogen carbon also forms a large number of compounds or more commonly called hydrocarbons due to its property called catenation here hydrogen wishes to gain one electron to obtain duplet atomic structure of helium its closest noble gas element and carbon requires eight electrons in its outermost shell to attain a stable electronic configuration therefore carbon give one electron to each of the four hydrogen atoms and hence hydrogen attain the duplet electronic configuration having two electrons in its outermost shell and carbon loses four electrons from its outermost shell and four hydrogen atoms gain one electron from carbon forming methane covalent bond

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your disposal, you have 5-mL and 10-mL transfer pipets and volumetric flasks of sizes 100, 250, 500, and 1000mL. Which of the following serial dilutions will give you the 200.0μM solution?

Answers

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

500mM = 500mM * ((1M)/(1000M))= 0.5M

Consider the following serial dilutions.

1)

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(500 mL)= 5 * 10^(-3)M

10 mL of this solution is diluted to 250 ml

M_(final)= (M_(2)V_(2))/(V_(final))= ((5 * 10^(-3)M)(10.0mL))/(250 mL)= 2 * 10^(-4)M

Convert μM :

2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M

Therefore, The dilution scheme will result in a 200μM solution.

2)

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(100 mL)= 2.5 * 10^(-2)M

10 mL of this solution is diluted to 1000 ml

M_(final)= (M_(2)V_(2))/(V_(final))= ((2.5 * 10^(-2)M)(10.0mL))/(1000 mL)= 2.5 * 10^(-4)M

Convert μM :

2.5 * 10^(-4)M = (2.5 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 250 \mu M

Therefore, The dilution scheme will not result in a 200μM solution.

3)

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(100 mL)= 0.05M

5 mL of this solution is diluted to 1000 ml

M_(final)= (M_(2)V_(2))/(V_(final))= ((0.05M)(5mL))/(1000 mL)= 0.25 * 10^(-4)M

Convert μM :

0.25 * 10^(-4)M = (0.25 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 25 \mu M

Therefore, The dilution scheme will not result in a 200μM solution.

4)

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5mL))/(250 mL)= 0.01M

10 mL of this solution is diluted to 500 ml

M_(final)= (M_(2)V_(2))/(V_(final))= ((0.01M)(10mL))/(500 mL)= 2 * 10^(-4)M

Convert μM :

2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M

Therefore, The dilution scheme will result in a 200μM solution.

5)

Dilute 10  mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(250 mL)= 0.02M

10 mL of this solution is diluted to 1000 ml

M_(final)= (M_(2)V_(2))/(V_(final))= ((0.02M)(10mL))/(1000 mL)= 2 * 10^(-4)M

Convert μM :

2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M

Therefore, The dilution scheme will result in a 200μM solution.

How does biodiversity help sustain a population in an area

Answers

Having a variety is always good you never want to have too much of one thing. It helps with the food chain its basically survival of the fittest 

Is it always possible to identify something as an element, compound, pure substance or mixture just by looking at it? Explain using examples from the tables. (table attached)

Answers

Using, the table given above, one can conclude that, IT IS NOT ALWAYS POSSIBLE TO IDENTIFY A SUBSTANCE AS A COMPOUND, MIXTURE, PURE SUBSTANCE OR AN ELEMENT JUST BY LOOKING AT THEM.

Looking at the table, you will observe that, some of the substances listed under mixtures are actually compounds. That is, combination of hydrogen and oxygen will form water, which is a compound. Also, combination of sodium and chlorine will form sodium chloride, which is a compound.

no, as some compounds(such as carbon dioxide and oxygen) are completely transparent, you can not see them and can not distinguish between the two with just the naked eye.

Magma that is low in silica has little __________ among its particles, which results in low viscosity.1.aa
2.pahoehoe
3.friction
4.viscosity
5.element
6.silicia
7.compound
8.carbon

Answers

Correct answer choice is:


3) Friction


Explanation:


Low viscosity relates to materials that are thin, such as water, while high viscosity materials are thick. The viscosity of a fluid is a sum of its opposition to progressive deformation by shear strain or tensile stress. For liquids, it compares to the simple idea of thickness, for example, honey has a very greater viscosity than water.

Answer:

Option (3)

Explanation:

The magma that is comprised of low silica content is termed as low viscosity magma. By low viscosity magma, it means that the magma can easily flow over the surface, without any difficulty. This indicates that there is less friction between the surface and the flowing magma. Friction usually refers to the resistant force that obstructs the flow of a liquid in any direction.

When there is a high frictional force, then the liquid flows very slowly with much difficulty. On the other hand, when there is a low force of friction, then the liquid can flow easily over a surface.

Thus, in a low viscosity magma, there is low silica content, so there is less frictional force between the surface and the magma, and the magma flows easily over the surface.

Hence, the correct answer is option (3).

What is the correct classification of a mixture in which both a solid and a liquid are visible?A.a homogeneous mixture and a solution
B.a homogeneous mixture and a suspension
C.a heterogeneous mixture and a solution
D.a heterogeneous mixture and a suspension

Answers

Answer is: D.a heterogeneous mixture and a suspension.

Suspension is a heterogeneous mixture (solute particles do not dissolve), that contains solid particles (for example nickel sulfide or NiS) sufficiently large for sedimentation.  

The internal phase (solid nickel sulfide) is dispersed throughout the external phase (water).

Another example of a suspension is sand in water.  

It would be D.a heterogeneous mixture and a suspension hope this helps