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ikadub [295]
3 years ago
11

What is food nutrients​

Chemistry
2 answers:
NikAS [45]3 years ago
7 0

Answer: In simple terms nutrients are the energy that you get from food certain foods give more nutrients and others give close to none. That is what nutrients in your food is

Explanation:

Illusion [34]3 years ago
5 0

Answer:

Nutrients arw compounds in foods essential to life and heath

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Which of the following statements correctly describes boyle's law
abruzzese [7]

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3 0
3 years ago
5. Chlorine gas was formed at the ... *
Marat540 [252]

Clorine gas was formed at the <em><u>positive</u></em><em><u> </u></em><em><u>electrode</u></em><em><u>.</u></em><em><u> </u></em>

5 0
2 years ago
How does soil structure affect the characteristics of soil? a. It determines the acidity of soil. b. It determines the available
7nadin3 [17]

Answer:

B

Explanation:

Soil structure affects water and air movement in a soil, nutrient availability for plants, root growth, and microorganism activity. ... This allows for greater air and water movement and better root growth.

4 0
3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to
AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength 6.32x10^{-36}m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

\lambda =\frac{h}{m*v}

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m

Best regards!

7 0
2 years ago
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