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2 years ago

Determine the overall charge on particles with the subatomic makeups given.

Chemistry

1 answer:

Gwar [14]2 years ago

5 0

Answer:

The correct solution will be "+3".

Explanation:

The given values are:

Number of protons

= 13

Number of neutrons

= 14

Number of electrons

= 10

As we know,

⇒ $Charge = Number \ of \ protons - Number \ of \ electrons$

On putting the estimated values, we get

⇒ $=13 - 10$

⇒ $=+3$

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Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

2 years ago

A branched alkane has ________ boiling point relative to the isomeric linear alkane. there are ________ london force interaction

olya-2409 [2.1K]

A branched alkane has HIGHER boiling point relative to the isomeric linear alkane. There are STRONGER london force interactions in the branched alkane.

:-) ;-)

4 0

2 years ago

When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m

Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

$500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2$

$42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl$

Have a nice day!

#LearnwithBrainly

7 0

2 years ago

Calculate the ratio of effusion rates of cl2 to f2 .

Lelechka [254]

Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)

4 0

2 years ago

Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.

vazorg [7]

The answer to this question is D! The ball and stick model! Hope this helps :)

8 0

2 years ago

Read 2 more answers