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Dominik [7]
3 years ago
12

Plant can accomplish something animal cells can’t. They can make their own food during

Chemistry
2 answers:
k0ka [10]3 years ago
8 0

Answer:

photosynethsis

Explanation:

Kruka [31]3 years ago
3 0

Answer:

Photosynthesis is the answer.

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
He generic metal a forms an insoluble salt ab(s) and a complex ac5(aq). the equilibrium concentrations in a solution of ac5 were
elena55 [62]
Balanced chemical reaction: A + 5C ⇄ AC₅.
<span>[A] = 0.100 M; equilibrium concentration.
</span><span>[C] = 0.0380 M.
</span>[AC₅] = 0.100 M.
Kf = [AC₅] / ([A] · [C]⁵).
Kf = 0.100 M ÷ (0.100 M · (0.0380 M)⁵.
Kf = 12620658.54 = 1,26·10⁷.
<span>The formation constant can be calculated when </span>chemical equilibrium is reached, when the forward reaction rate is equal to the reverse reaction rate.
7 0
3 years ago
A 4.5-liter sample of a gas has 0.80 mole of the gas. If 0.35 mole of the gas is added, what is the final volume of the gas? Tem
AVprozaik [17]

Answer:

6.5 liters will be your answer

Explanation:

5 0
3 years ago
Read 2 more answers
What must a plant need to produce 4 molecules of sugar?
Amiraneli [1.4K]

Answer:

24 molecules of H20

Explanation:

Photosynthesis:

It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.

Carbon dioxide + water + energy →   glucose + oxygen

water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture  by chloroplast.

Chemical equation:

6H₂O + 6CO₂ + energy  →   C₆H₁₂O₆ + 6O₂

it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.

consider 1 mole contain 1 molecule.

So in order to produced 4 molecules of sugar number of molecules of water carbon dioxide needed are,

6 × 4 = 24 molecules

24 molecules of water and 24 molecules of carbon dioxide needed.

4 0
3 years ago
How many grams of mgcl2 is produced if 4. 67 moles of hcl react
frosja888 [35]

The mass of MgCl₂ produced is 5.685 grams.

<h3>What is mass?</h3>

Mass is the quantity of matter of a physical body.

The reaction is

Mg   +   2HCl   →    MgCl₂ + H₂

2 mole of HCl

(2 × 36.458) = 72.8 g        

                   

1 moles of MgCl₂

(1 × 95.21 g) = 95.21 g

1 g  Mg will produce = \dfrac{95. 21}{78.2}

4.67 mol Mg will produce

\dfrac{95. 21}{78.2} \times 4.67 = 5.685

Thus, the mass of MgCl₂ produced is 5.685 grams.

Learn more about mass

brainly.com/question/19694949

#SPJ4

7 0
2 years ago
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