All categories

3 years ago

Please help brainiest is award!!!!!!!!!!!

Chemistry

2 answers:

Alekssandra [29.7K]3 years ago

7 0

Answer:

it doesnt even give you enough information for you to answer it

Explanation:

slega [8]3 years ago

5 0

Answer:

it doesnt even give you enough information for you to answer it

Explanation:

You might be interested in

How many grams of sucrose (c12h22o11) are in 1.55 l of 0.758 m sucrose solution? express your answer in grams to three significa

satela [25.4K]

The molecular formula of sucrose is - C₁₂H₂₂O₁₁
molecular mass of sucrose - 342 g/mol
molarity of sucrose solution is 0.758 M
In 1 L solution the number of sucrose moles are - 0.758 mol
Therefore in 1.55 L solution, sucrose moles are - 0.758 mol/L x 1.55 L
= 1.17 mol
The mass of 1.17 mol of sucrose is - 1.17 mol x 342 g/mol = 4.00 x 10² g

7 0

3 years ago

Read 2 more answers

Which of the following can’t be separated into simpler substances or converted into another substance by chemical process?

timofeeve [1]

Your answer would be an Element can't be separated into simpler substances or converted into another substance by chemical process.

Hope that helps!!!

8 0

3 years ago

Which of the following will produce a precipitate?

nekit [7.7K]

Reaction of option c produces precipitate.

Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.

Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.

Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:

$Co^2^+ + Zn(NO¬_3 )¬_2 Co(NO¬_3 )¬_2 + Zn^2^+$

Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution

8 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

A form of energy produced my a machine

sergij07 [2.7K]

Sound, heat, light etc

3 0

3 years ago