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3 years ago

How many liters of Cl2 gas will you have if you are using 63 g of Na?

Chemistry

1 answer:

ELEN [110]3 years ago

6 0

Answer:

You will have 19.9L of Cl2

Explanation:

We can solve this question using:

PV = nRT; V = nRT/P

Where V is the volume of the gas

n the moles of Cl2

R is gas constant = 0.082atmL/molK

T is 273.15K assuming STP conditions

P is 1atm at STP

The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:

63g * (1mol / 70.906g) = 0.8885 moles

Replacing:

V = 0.8885mol*0.082atmL/molK*273.15K/1atm

V = You will have 19.9L of Cl2

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In two or more complete sentences explain how to balance the chemical equation, KClO3 ⟶ KCl + O2 and include all steps. Write yo

Serggg [28]

Answer:

2KClO3 -------> 2KCl + 3O2

Explanation:

First, in balancing a chemical reaction such as the one given in the question, you should understand that for an equation to be balanced, the number of atoms and ions on both sides of the equation that is the right and left side must be equal. This follows the law of conservation of mass which tells us that matter can neither be created nor destroyed but can be changed into another form.

Next is to begin balancing the equation by identifying and writing down the substances given:

KCl03 ---------> KCl + O2

Next is to count he number of the individual atoms on each side and find out if they are the same on both sides and if not you must follow the next step.

Add a corresponding number and use it to multiply the atoms involved

KClO3 ---------> KCl + O2

Oxygen is 3 on the left side and two on the other side, so we multiply the left hand side by 2 and the right hand side by 3

2KClO3 -----> KCl + 3O2

The potassium and Chlorine are no longer balanced, so you multiply the right had=nd side of KCl by 2.

2KClO3 -----> 2KCl + 3O2

The reactionis herefore balanced as both sides have equal number of atoms and ions.

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4 years ago

What is hypothesis please help

Mama L [17]

A hypothesis is how you think the experiment is going to end

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3 years ago

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A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible

RideAnS [48]

Answer:

TS (total solid, mg/L) = 57,174 mg/L

VS (volatile solid, mg/L) = 24,088 mg/L

Explanation:

First step to solve this problem is to know data and questions:

Data:

Sample volume = 50 mL

Sample weight = 50.1 g

P1 (weight empty crucible) = 17.1234 g

P2 (weight after drying a 103º in oven) = 19.9821 g

P3 (weight after incinatrion 550º) = 18.7777 g

Questions:

TS = ?

VS = ?

Formula:

We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:

$TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}$

$VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}$

We have to transform mL to L so we will divide mL by 1,000 the sample volume:

$Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L$

In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)

Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:

$TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L$

$VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L$

We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.

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3 years ago

To begin the experiment, 1.5g of methane CH4 is burned in a bomb calorimeter containing 1,000 grams of water. The initial temper

vredina [299]

Mass of methane takne = 1.5g

moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles

mass of water = 1000 g

Initial temperature of water = 25 C

final temperature = 37 C

specific heat of water = 4.184 J /g C

1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules

2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J

3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules

This heat is released by 0.094 moles of methane

So heat released by one mole of methane =

- 622851.06 Joules = 622.85 kJ / mole

4) standard enthalpy of combustion = -882 kJ / mole

Error = (882-622.85) X 100 / 882 = 24.84 %

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