Increasing the concentration of a reactant speeds up a reaction.<br> True<br> False

175 mL of Cl2 gas is held in a flexible vessel at

Gnoma [55]

Answer:

V₂ = 15.6 L

Explanation:

Given data:

Initial volume = 175 mL (0.175 L)

Initial pressure = 1 atm

Initial temperature = 273 K

Final temperature = -5°C (-5+273 = 268 K)

Final volume = ?

Final pressure = 1.16 kpa (1.16/101=0.011 atm)

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm

V₂ = 46.9 L / 3.003

V₂ = 15.6 L

7 0

2 years ago

Read 2 more answers

Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be

Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

HCOOH (formic acid) → HCOO⁻ (formate) + H⁺

H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺

Formal Charges:

Formal charges are calculated using following formula,

F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

F.C = [6] - [6 + 2/2]

F.C = [6] - [6 + 1]

F.C = 6 - 7

F.C = -1

For Sodium:

F.C = [1] - [0 + 0/2]

F.C = [1] - [0]

F.C = 1 - 0

F.C = +1

5 0

2 years ago

6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021

Aleks [24]

Answer:

The total amount of Au is $ $2.0\times10^{24}$

Explanation:

Given that,

Mass of 1.0 ml of Au $m=4\times10^{-2}\ g$

Total volume of water in oceans $V=1.5\times10^{21}\ L$

We need to calculate the volume in ml

Using given volume

$V=1.5\times10^{21}\times1000\ mL$

$V=1.5\times10^{24}\ mL$

We need to calculate the total mass of Au

Using given data

$1\ ml\ volume = 4\times10^{-2}\ g$

$1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}$

So, The total mass of Au is $6\times10^{22}\ g$

The mass will be in ounce,

$Mass=0.035274\times6\times10^{22}$

$Mass=2.12\times10^{21}\ ounce$

The total amount of the Au Will be

$Total\ amount=2.12\times10^{21}\times948$

$Total\ amount=2.0\times10^{24}$

Hence, The total amount of Au is $ $2.0\times10^{24}$

3 0

1 year ago

This information is taken directly from Exercise 3 in the CHEM 111/112 Laboratory Manual. Another example of excellence in the p

kolbaska11 [484]

Answer:

the information was not cited correctly....

Explanation:

I hope the following explanation will help you a lot.

5 0

2 years ago

Which traits does the common garter snake have that might be adaptive for the environment where it lives?

Licemer1 [7]

Answer:

As the garter snake can be found almost in any kind of habitat, what makes them be able to survive in any environment include:

1. They hibernate to increase their chances of survival in unfavorable weather conditions.

2. They can blend with the background of any environment especially grass to escape being eaten.

3. They produce an odor that is usually unpleasant especially when about to be attacked.

Explanation:

The garter snakes are distinguished by the three stripes running the length of their body and can often be found in forests, places that are even close to water bodies, and almost any place, even in holes.

4 0

1 year ago

Increasing the concentration of a reactant speeds up a reaction. True False