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3 years ago

How would one go about solving this?? Thank you.

Chemistry

1 answer:

Marta_Voda [28]3 years ago

7 0

Answer:

5.66g/ml

Explanation:

Mass of Metal Slug, m = 25.17g

Volume of container (flask), v = 59.7ml

Mass of Methanol, M = 43.7g with density, d = 0.791g/ml

Therefore,

Volume of Methanol = M/d = 43.7 / 0.791 = 55.25ml

Volume of Metal slug in the flask = v - V = 59.7 - 55.25 = 4.45ml

Density of Metal slug = mass of metal slug / volume of metal slug

= 25.17 / 4.45 = 5.66g/ml

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What is a cut circuit

Luden [163]

Answer:

a CUT stands for Circuit Under Test.

Explanation:

6 0

2 years ago

What is the mass in grams of 14.2 liters of sulfur dioxide gas at STP?

ICE Princess25 [194]

Answer:

Mass = 40.4 g

Explanation:

Given data:

Mass in gram = ?

Volume of SO₂ = 14.2 L

Temperature = standard = 273 K

Pressure = standard = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

1 atm × 14.2 L = n × 0.0821 atm.L/ mol.K × 273 K

14.2 atm.L = n × 22.41 atm.L/ mol

n = 14.2 atm.L/22.41 atm.L/ mol

n = 0.63 mol

Mass of sulfur dioxide:

Mass = number of moles × molar mass

Mass = 0.63 mol × 64.1 g/mol

Mass = 40.4 g

7 0

2 years ago

For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r

s344n2d4d5 [400]

Answer:

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

3 0

3 years ago

The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from

balu736 [363]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 100 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 600 mm of Hg

Final pressure ( $P_{2}$ ) = 1200 mm of Hg

Initial volume ( $V_{1}$ ) = 200 ml

To find:

Final volume ( $V_{2}$ )

We know;

According to the ideal gas equation,

P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

From the above mentioned equation,

P × V = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{V_{1} }{V_{2} }$

Where,

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

So;

$\frac{600}{1200}$ = $\frac{V_{2} }{200}$

$V_{2}$ = 100 ml

Therefore the final volume of the gas is 100 ml.

5 0

3 years ago

What would be the mass of 9.03*10^21 molecules of hydrobromic acid

Fynjy0 [20]

Answer:

2.11 g hydrobromic acid (correct to 3SF)

Explanation:

Molecular formula of hydrobromic acid = C2H5BrO2

mass of C2H5BrO2 = 140.96g

Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.

It looks like this:

9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2

3 0

3 years ago

How would one go about solving this?? Thank you. ​

How would one go about solving this?? Thank you.