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3 years ago

What is the active component of dynamite

Chemistry

1 answer:

yawa3891 [41]3 years ago

4 0

The active component in dynamite is Nitroglycerin, and it is in most of the other explosive material.
Hoped I helped:D

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A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68

FinnZ [79.3K]

First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.

P $_{T}$ = total pressure

P $_{Ne}$ = pressure of neon

P $_{Ar}$ = pressure of argon

P $_{He}$ = pressure of helium {which is required}

P $_{T}$ = P $_{Ne}$ + P $_{Ar}$ + P $_{He}$

1.25 = 0.68 + 0.35 + P $_{He}$

P $_{He}$ = 1.25 - [0.68 + 0.35] = 0.22 atm

4 0

3 years ago

a. If 42.5 g of CH3OH reacts with 22.8 L of O2 at 27°C and a pressure of 2.00 atm, calculate the number of grams of water vapor

Korvikt [17]

Answer:

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

Explanation:

Step 1: Data given

Mass of CH3OH =42.5 grams

Molar mass CH3OH = 32.04 g/mol

Volume of O2 = 22.8 L

Pressure = 2.00 atm

Step 2: The balanced equation

2CH3OH + 3O2 → 2CO2 + 4H2O

Step 3: Calculate moles CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 42.5 grams / 32.04 g/mol

Moles CH3OH = 1.326 moles

Step 4: Calculate moles O2

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of O2 = 22.8 L

⇒with n = the moles of O2 = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

n = (p*V) / (R*T)

n = (2.00 * 22.8) / (0.08206*300)

n = 1.85 moles

Step 5: Calculate the limiting reactant

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

O2 is the limiting reactant. It will completely be consumed ( 1.85 moles). CH3OH is in excess. There will react 2/3*1.85 = 1.233 moles. There will remain 1.326 - 1.233 = 0.093 moles

Step 6: Calculate moles products

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 H2O

For 1.85 moles O2 we'll have 1.233 moles CO2 and 2.467 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.467 moles * 18.02 g/mol

Mass H2O = 44.46 grams

Step 8: Calculate volume H2O

p*V = n*R*T

⇒with p = the pressure = 2.00 atm

⇒with V = the volume of H2O = TO BE DETERMINED

⇒with n = the moles of H2O = 2.467 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 27 °C = 300 K

V = (n*R*T)/p

V = (2.467 * 0.08206 * 300) / 2.00

V = 30.37 L

The mass of water vapor is 44.46 grams

The volume of water is 30.37 L

3 0

3 years ago

Can someone pls answer quickly?

liq [111]

answer is atom.Hope it is helpful.

3 0

3 years ago

Read 2 more answers

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 0

3 years ago

4. Explain error and how it can impact your results from an experiment

Jet001 [13]

Random errors will shift each measurement from its true value by a random amount and in a random direction. These will affect reliability (since they're random) but may not affect the overall accuracy of a result.

3 0

3 years ago