Is berkelium flammable

Calculate the mass percent for all components in a solution containing the following. 0.350 Kg of water, 5.4 moles of ammonia an

Alina [70]

Answer:

% Water = 54.96%

% Ammonia = 0.14%

% Cobalt (II) Nitrate = 30.62%

Explanation:

To calculate mass percent, first we need to calculate the total mass of the mixture:

Mass Water ⇒ 0.350 kg Water = 350 g water

Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol = 91.8 g

Mass cobalt (II) nitrate ⇒ 195.0 g

Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate

Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g

To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:

% Water ⇒ 350/636.8 * 100% = 54.96%

% Ammonia ⇒ 91.8/636.8 * 100% = 0.14%

% Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%

8 0

3 years ago

Determine the identity of a cube of metal that measures 1.2 cm on each side and has a mass of 15.4g.

FromTheMoon [43]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
density 15.4 grams per 1.2³ cm³ ≅ 8.9 grams per cm³
Find the metal that has a density of approximately 8.9 g/cm³

8 0

3 years ago

Read 2 more answers

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

3 years ago

“Gasoline boils at a relatively low temperature (about 150°C). The kerosene is removed at around 200°C, followed by diesel oil a

ANEK [815]

Well its Organic Chemistry... and is the Fractional Distillation of Crude Oil

3 0

3 years ago

Table salt is a refined salt containing about 97 to 99% sodium Chloride iodized salt containing potassium iodide is widely avail

Free_Kalibri [48]

Answer:

Explanation:

Sodium chloride is ionic compound. It is formed by the transfer of electron from one atom to the atom of another element.

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are bonded together electrostatic attraction occur between anion and cations.

Sodium atom have one valance electron by losing this one valance electron sodium atom get the complete octet. Chlorine atom has seven valance electrons and needed to lose seven valance electrons or to get one electron and thus complete the octet. It is very easy for chlorine atom to get one electrons instead of losing all seven electron. Thus when it react with sodium it gain the valance electron of sodium and form ionic compound.

That's why only one atom of sodium combine with one atom chlorine.

8 0

3 years ago