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3 years ago

Which of the following are common decomposers? (Select all that apply) fungi, ferns, bacteria, viruses, grasses

Chemistry

2 answers:

elena-14-01-66 [18.8K]3 years ago

5 0

Answer:

Fungi, bacteria, ferns,

Explanation:

bacteria Fungi Ferns are important. decomposers,

I hope it helps you.

antiseptic1488 [7]3 years ago

4 0

Fungi heijsiwhqibwudyenejidjd

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I need help with this question please!!

mariarad [96]

The answer is B

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3 0

3 years ago

What is the median in the box plot? 50 60 ТО 80 90 100

d1i1m1o1n [39]

Answer:

60 to 80

<h3>Explanation:</h3><h3> Because the interquartile range IQR</h3>

Q1 IS 25TH Percentile the Q3 75th percentile

3 0

3 years ago

Read 2 more answers

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What

ryzh [129]

Answer : The mass of $PbI_2$ precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of $PbI_2$ = 461.01 g/mole

First we have to calculate the moles of $NaI$ .

$\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles$

Now we have to calculate the moles of $PbI_2$ .

The balanced chemical reaction is,

$Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaI$ react to give 1 mole of $PbI_2$

So, 0.042 moles of $NaI$ react to give $\frac{0.042}{2}=0.021$ moles of $PbI_2$

Now we have to calculate the mass of $PbI_2$ .

$\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2$

$\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g$

Therefore, the mass of $PbI_2$ precipitate produced will be, 9.681 grams.

6 0

4 years ago

What is the molarity of a solution that contains 0.400 mol HCl in 9.79 l solution?

Maslowich

Answer:

Option B, 0.0409 M

Explanation:

1) Data:

a) M = ?

b) n = 0.400 mol

V = 9.79 liters

2) Formula:

M = n / V (in liters)

3) Solution:

M = 0.400 mol / 9.79 liter = 0.04086 M

Since each data contain 3 significant digits, the answer must be reported with 3 significant digits. So, the correct answer is 0.0409 M, which is the option B.

8 0

4 years ago

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0

3 years ago

Which of the following are common decomposers? (Select all that apply) fungi, ferns, bacteria, viruses, grasses​

Which of the following are common decomposers? (Select all that apply) fungi, ferns, bacteria, viruses, grasses