Answer:

Explanation:
Density can be found by dividing the mass by the volume.

The mass of the water is 50.0 grams.
The volume of the water is 50.0 milliliters.

Substitute the values into the formula.

Divide.

The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³
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<span>The Mechanical energy in a mechanical system is determined by adding the potential and kinetic energy together. </span>
Answer:
³⁸₂₀Ca.
Explanation:
³⁸₁₉K –> __ + ⁰₋₁β
Let ʸₓA represent the unknown.
Thus the equation above can be written as:
³⁸₁₉K –> ʸₓA + ⁰₋₁β
Thus, we can obtain the value of y an x as follow:
38 = y + 0
y = 38
19 = x + (–1)
19 = x – 1
Collect like terms
19 + 1 = x
x = 20
Thus,
ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca
Therefore, the equation is:
³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β
Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature =
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 = 
C = 0.0285
This also means that,
= 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
= 
Now, let us assume that mass of
= x grams
And, mass of
= (1.00 - x)
So, moles of
=
Now, moles of
=
=
= x = 0.346
Therefore, we can conclude that amount of
present is 0.346 g and amount of
present is (1 - 0.346) g = 0.654 g.
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.