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3 years ago

What would happen if Phosphorus disappeared?

1 answer:

3 0

The cycles would eventually come to an end. If the sun were to burn out, the Earth would lose the energy and heat from the sun and soon lose its own heat and energy and fall apart into space. The magnetism would soon become demagnetized, plates would stop moving, trees would die, life etc. no more O2, CO2, cycle and all others would fade.

You might be interested in

How do you find the pressure in a container if you don't have the temperature?

BARSIC [14]

It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
PV = constant or P1V1 = P2V2 where P1V1 are initial conditions and P2V2 are final.

For adiabatic process,

PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.

where gamma = Cp
------
Cv
Cp = specific heat at constant pressure and Cv = specific at constant volume.

Value of Gamma will be given in question.
Hope this helps!

7 0

3 years ago

A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n

Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution) × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

#SPJ4

8 0

1 year ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

5 0

3 years ago

How many miles of glucose are in 19.1 g of glucose

geniusboy [140]

Answer:

did you mean moles? If so, answer is down below.

Explanation:

there are 0.106 moles of glucose in 19.1 g of glucose.

7 0

3 years ago

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0

3 years ago