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3 years ago

Which reason best explains why metals are shiny?

Chemistry

1 answer:

alexgriva [62]3 years ago

8 0

Reflection, refracting, and the energy levels of molecular orbitals

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If air contains 78% Nitrogen by volume, what is the solubility of nitrogen in water at 25 degree celisus

Svet_ta [14]

The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L

<h3>Further explanation</h3>

Given

78% Nitrogen by volume

Required

The solubility of nitrogen in water

Solution

Henry's Law states that the solubility of a gas is proportional to its partial pressure

Can be formulated

S = kH. P.

S = gas solubility, mol / L

kH = Henry constant, mol / L.atm

P = partial gas pressure

In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:

Vn / Vtot = Pn / Ptot

78/100 = Pn / 1

Pn = 0.78 atm

Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm

The solubility :

$\tt S=6.25.10^{-4}\times 0.78\\\\S=4.88\times 10^{-4}~mol/L$

7 0

2 years ago

there are 125 lima beans 56 garbanzo beans 39 kidney beans and 144 green peas in a container. find the percent by number of gree

Sauron [17]

39.6% of the beans are green beans

5 0

3 years ago

Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

Gwar [14]

<u>Answer:</u>

<em>0.264 g of $CO_2$ can be formed from 288 mg of $O_2$ </em>

<u>Explanation:</u>

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

Moles $O_2$ is converted to moles $CO_2$ by using the mole ratio of $O_2:CO_2$ is 9 : 6

Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

=0.264g (Answer)

7 0

3 years ago

If we react 5.4g of sodium chloride with an unknown amount of fluorine gas, we produce 4.9g of sodium fluoride and 3.7g chlorine

Bumek [7]

Answer:

4.43 g

Explanation:

The reaction between sodium chloride and flourine gas is given as;

NaCl + F2 --> NaF + Cl2

From the stochiometry of the equation;

1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2

Mass of 1 mol of F2 = 38g

Mass of 1 mol of sodium flouride, NaF = 42g

This means 38g of flourine reacted with NaCl to form 42g of NaF

xg of F2 would form 4.9g of NaF

38 = 42

x = 4.9

x = 4.9 * 38 / 42

x = 4.43 g

4 0

3 years ago

Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes. How will the new soles h

HACTEHA [7]

Maybe her old shoes had soft worn out bottoms and she slips in them. So her new shoes had more grip than her old ones so they kept her from falling.

6 0

3 years ago

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