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3 years ago

Please help I don’t know if my answer is right!!

Chemistry

1 answer:

Temka [501]3 years ago

8 0

Answer:

neutralization reaction! Aka: option C!

HOPE THIS HELPS! :)

Explanation:

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sergey [27]

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3 years ago

Сколько граммов соли образуется, если в раствор, содержащий 9,8 г серной кислоты, добавим другой раствор, имеющего 10 г гидрокси

lutik1710 [3]

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3 years ago

What do scientists Need to know about a compound in order to Classify it

OleMash [197]

Compounds are classified according to the elements that make them up. For example, oxides contain one or more oxygen atoms, hydrides contain one or more hydrogen atoms.
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3 years ago

Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s

stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 = 0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x = 1.053 g / 0.0405 mol = 26 g/mol

7 0

3 years ago

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0

3 years ago