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3 years ago

If you shake up different kinds or brands of soft drinks will they all spew you the same amount

Chemistry

2 answers:

Morgarella [4.7K]3 years ago

3 0

Answer:

yeah and the point?

Explanation:

Harman [31]3 years ago

3 0

Answer:

The different brands of sodas will not spew the same amount when shaken because there are different amounts of sugar in the sodas and that will affect the amount spewing. Pepsi will spew the highest, compared to the other brands of sodas

Explanation:

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15 points!

Alex_Xolod [135]

Answer:

C: It is the difference between the total kinetic and potential energy in a system.

Explanation:

6 0

3 years ago

Please help!! i need the answer ASAP!

nirvana33 [79]

Answer:

23.34 %.

Explanation:

The percentage of water must be calculated as a mass percent.

We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.

Atomic masses of the elements:

Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.

Mass of the formula unit:

Cd(NO₃)₂•4H₂O

mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.

Mass of water in the formula unit:

mass of water = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.

So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %

7 0

3 years ago

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

3 years ago

Briefly explain how magnetic behavior is related to the arrangement of valence electrons in the outer orbitals

olga nikolaevna [1]

Answer:

Elements with unpaired valence electrons in the outer orbital are paramagnetic. Elements that are predicted to exhibit paramagnetic behavior may exhibit ferromagnetic behavior. Elements that contain only paired valence electrons are diamagnetic.

3 0

4 years ago

Consider the following exothermic reaction and predict how the change below will affect the concentration (increase, decrease, s

aev [14]

Answer:

a) equilibrium shifts towards the left

b) equilibrium shifts towards the right hand side.

c)equilibrium shifts towards the left hand side

d) addition of argon has no effect on the equilibrium position

e) equilibrium position shifts towards the left hand side

f) equilibrium position is shifted towards the right hand side

g) addition of a catalyst has no effect on the equilibrium position

Explanation:

Given the equation;

3N2H4(g)⇄ 4NH3 (g) + N2 (g)

Adding N2

The addition of N2 will increase the concentration of N2 in the system thereby shifting the equilibrium position to the left in accordance with Le Chatelier's principle.

Removing N2

The removal of N2 drives the forward reaction and the equilibrium shifts towards the right yielding more N2 in accordance with Le Chateliers principle.

Add NH3

The addition of NH3 will shift the equilibrium position towards the left hand side according to Le Chateliers principle.

d) addition of argon has no effect on the equilibrium position.

e) increasing the temperature

Since the reaction is exothermic, increasing the temperature favours the reverse reaction and the equilibrium position his shifted towards the left hand side.

f) Decrease in volume;

Decreasing the volume favours the forward reaction hence the equilibrium position is shifted towards the right.

g) addition of a catalyst has no effect on the equilibrium position.

3 0

3 years ago