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3 years ago

Explain in your own words how DDT moved through the environment starting with spraying the farm crops

Chemistry

1 answer:

S_A_V [24]3 years ago

7 0

Answer:

here and a cylinder have the same radius and height. The volume of the cylinder is .

What is the volume of the sphere?

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If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

Luba_88 [7]

<u>Answer:</u> The value of $K_c$ is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

The expression for the $K_{c}$ is given as:

$K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

We are given:

$[NH_3]=0.25M$

$[H_2]=0.75M$

$[N_2]=1.1M$

Putting values in above equation, we get:

$K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}$

$K_c=0.135$

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the given reaction, the value of $K_c$ is less than 1. Thus, the reaction is reactant favored.

Hence, the value of $K_c$ is 0.136 and is reactant favored.

4 0

3 years ago

Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3

Vadim26 [7]

Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g

7 0

3 years ago

Read 2 more answers

5. Why are the noble gases such as helium (He) and neon (Ne) grouped

GalinKa [24]

Answer:

A. They are nonreactive gases

8 0

2 years ago

If one gram of sulphur dioxide contains x molecules what will be the number of molecules in 1g of methane

evablogger [386]

The ratio of molecules in sulphur dioxide and methane will be the same as the ratio of their moles. So, first of all we should find out the number of moles of sulphur dioxide in 1 gram of sulphur dioxide in 1 gram of sulphur dioxide, and the number of moles of methane in 1 gram of methane. This can be done as follows :

(i) The molecular formula of sulphur dioxide is $SO_{2}$

So, $1$ mole of $SO_{2}$ = $Mass$ $of$ $2'O'$

$=32+2*16$

$= 64$ grams

Now, $64g$ of sulphur dioxide $= 1$ mole

So, $1g$ of sulphur dioxide = $\frac{1}{64}$ mole

Thus, we have $\frac{1}{64}$ mole of sulphur dioxide and it contains molecules in it. Now, since equal moles of all the substance contain equal number of molecules, therefore, $\frac{1}{64}$ mole of methane will also contain x molecules of methane.

(ii) Molecular formula of methan is $CH_{4}$