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3 years ago

Convert 100 °C into Kelvin

Chemistry

1 answer:

Mice21 [21]3 years ago

6 0

Hey there mate ;)!,

Given : 100°C

To convert to Kelvin scale unit,

Add the Celsius unit with 273.

→ (100+273) K

→ 373 K

Therefore the answer is 373 Kelvin or 373 K.

By Benjemin ☺️

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Which option below describes a chemical property of a substance?

boyakko [2]

Answer:

Explanation:

Rust is a chemical change (reaction) all of the other options only change the state meaning that it is a physical reaction.

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3 years ago

Read 2 more answers

The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million ti

klasskru [66]

Answer:

54.9 kJ/mol

Explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

$k=A.e^{-Ea/RT}$

where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So $k_{B}=10^{6} k_{A}$ .

Then,

$k_{B}=10^{6} k_{A}\\A.e^{-Ea_{B}/RT}=10^{6}A.e^{-Ea_{A}/RT}\\e^{-Ea_{B}/RT}=10^{6}e^{-Ea_{A}/RT}\\ln(e^{-Ea_{B}/RT})=ln(10^{6}e^{-Ea_{A}/RT})\\\frac{-Ea_{B}}{RT} =ln10^{6} -\frac{Ea_{A}}{RT} \\Ea_{B}=(ln10^{6} -\frac{Ea_{A}}{RT}).(-RT)=(ln10^{6}-\frac{89.1kJ/mol}{(8.314\times 10^{-3} kJ/mol.K).298K} ).(-8.314\times 10^{-3} \frac{kJ}{mol.K}.298K )=54.9kJ/mol$

8 0

3 years ago

explain why metals conduct electricity in their solid forms but ionic compounds do not. you may use a picture to support your an

deff fn [24]

Explanation:

Ionic solids are defined as the solids in which atoms of opposite charge are held together by strong forces of attraction. As a result, in solid state these atoms are not present in free state.

Hence, they cannot move as ions from one place to another. On the other hand, metals are able to conduct electricity because of the presence of free electrons in it.

As we know that electricity is the flow of ions or electrons from one place to another. Therefore, metals are able to conduct electricity in their solid state.

7 0

3 years ago

A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point

Ksivusya [100]

Answer:

$P_{T}$ = 2.94 atm

Explanation:

The total pressure ( $P_{T}$ ) in the container is given by:

$P_{T} = P_{O_{2}} + P_{He}$

The pressure of the oxygen ( $P_{O_{2}}$ ) and the pressure of the helium ( $P_{He}$ ) can be calculated using the ideal gas law:

$PV = nRT$

Where:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

$n_{O_{2}} = \frac{m}{M}$

Where m is the mass of the gas and M is the molar mass

$n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles$

And the number of moles of helium is:

$n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles$

Now, we can find the pressure of the oxygen and the pressure of the helium:

$P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm$

$P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm$

Finally, the total pressure in the container is:

$P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm$

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!

6 0

4 years ago

Perform the following for Part C of this lab:

kaheart [24]

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

Titrant (NaOH) concentration: 1.0 M

Vinegar volume: 2.0 mL

Initial buret reading (initial NaOH volume): 0.1 mL

Final buret reading (final NaOH volume): 11.1 mL

a. Calculate the volume of NaOH that was added to the vinegar. Convert this volume to liters. Show your work.

The volume of NaOH is the difference between the final and the initial buret reading.

11.1 mL - 0.1 mL = 11.0 mL × (1 L/1000 mL) = 0.0110 L

b. Convert the measured volume of vinegar to liters. Show your work.

2.0 mL × (1 L/1000 mL) = 0.0020 L

c. Calculate the moles of NaOH using the volume and molarity of NaOH. Show your work. moles = molarity x volume

moles = molarity × volume

moles = (1.0 mol/L) × 0.0110 L = 0.011 mol

d. Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar is equal to the moles of NaOH reacted during the titration. Calculate the molarity of the acetic acid in the vinegar. Show your work. molarity = moles / volume

molarity = moles / volume

molarity = 0.011 mol/0.0020 L = 5.5 M

e. Calculate the grams of acetic acid in the vinegar. Show your work. mass = moles x molar mass (g/mol)

mass = moles × molar mass

mass = 0.011 mol × 60.05 g/mol = 0.66 g

f. Assuming that the density of vinegar is very close to 1.0 g/mL, the 2.0 mL sample of vinegar used in the titration should weigh 2.0 g. Use this to calculate the mass % of acetic acid in the vinegar sample. mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%

6 0

3 years ago