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4 years ago

Three synonyms for introduced species

Chemistry

1 answer:

maks197457 [2]4 years ago

6 0

Alien pest species, Harmful pest species, harmful non-indigenous species

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How to do my homework? i’ll give 20 points to anyone who will do it!

tensa zangetsu [6.8K]

what is it ??? ill edit this

3 0

3 years ago

Read 2 more answers

A ballon is inflated with 2.42L of helium at a temperature of 27.0°C. When put in the freezer, the volume changes to 2.37L and -

natita [175]

Answer:

838 torr

Step-by-step explanation:

To solve this problem, we can use the Combined Gas Laws:

p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁

p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁

p₁ = p₂ × V₂/V₁ × T₁/T₂

Data:

p₁ = ?; V₁ = 2.42 L; T₁ = 27.0 °C

p₂ = 754 torr; V₂ = 2.37 L; T₂ = -8.8 °C

Calculations:

(a) Convert temperatures to kelvins

T₁ = (27.0 + 273.15) K = 300.15 K

T₂ = (-8.8 + 273.15) K = 264.35 K

(b) Calculate the pressure

p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)

p₁ = 754 torr × 0.979 × 1.135

p₁ = 838 torr

5 0

4 years ago

The molar solubilities of the following compounds (in mol/L) are:

IceJOKER [234]

Answer: The decreasing order of $K_{sp}$ is $AgSCN>AgBr>AgCN$

Explanation:

For AgBr:

The balanced equilibrium reaction for the ionization of silver bromide follows:

$AgBr\rightleftharpoons Ag^{+}+Br^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][Br^-]$

We are given:

$s=7.3\times 10^{-7}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}$

Solubility product of AgBr = $5.33\times 10^{-13}$

For AgCN:

The balanced equilibrium reaction for the ionization of silver cyanide follows:

$AgCN\rightleftharpoons Ag^{+}+CN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][CN^-]$

We are given:

$s=7.7\times 10^{-9}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}$

Solubility product of AgCN = $5.33\times 10^{-17}$

For AgSCN:

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

$AgSCN\rightleftharpoons Ag^{+}+SCN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][SCN^-]$

We are given:

$s=1.0\times 10^{-6}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}$

Solubility product of AgSCN = $1.0\times 10^{-12}$

The decreasing order of $K_{sp}$ follows:

$AgSCN>AgBr>AgCN$

4 0

4 years ago

I will give you a brainliest.

Zina [86]

answer= 0.912 L or 912 mL

M(KClO3) = 122.55 g/mol

3.00 g KClO3 * 1 mol/122.55 g = 3.00/122.55 mol =0.02449 mol

2KCIO3(s)=2KCI(s) + 3O2(g)

from reaction 2 mol 3 mol

given 0.02449 mol x

x = 0.02449*3/2 =0.03673 mol O2

T = 24 + 273.15 = 297.15 K

PV = nRT

V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =

= 0.912 L or 912 mL

8 0

3 years ago

How many grams of O2 are produced as 0.033 mol of water decompose

PolarNik [594]

Answer: This can be quickly solved with "traintracks"

Explanation:

You start w/ grams of water and want to find moles of oxygen gas produced.

So you want to Convert:

Grams of water -> moles of water -> moles of oxygen gas.

The two things you need to know to set up the tracks are:

1)Molar mass of water- H2O

Hydrogen - 1.008(x2)

Oxygen - 16.00

Water - 18.016

6 0

3 years ago