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3 years ago

Wirte the benefits and harmful effects of micro organisms in our life? ans

Chemistry

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

Microorganisms have uses and benefits across all aspects of human life. From the bacteria that help humans digest food to the viruses that help plants resist heat, bacteria, viruses and fungi – when used properly – are key components in food, medicine, agriculture and other areas. In the future, they may even be core components of infrastructure and other new technologies.

Explanation:

You mean microorganisms, yes?

You might be interested in

What is the pH of pure water at 40.0Â°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233

Sergeeva-Olga [200]

Answer:

PH= 6.767 (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

$[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\$

at 40ºC

$[H_{3}O^{+} ] = [OH^{-} ]$

$[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}$

$[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M$

$PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767$

7 0

3 years ago

Identify the only polyatomic ion below that is NOT an oxyanion. Your answer: A Cyanide B Nitrate C Hydroxide D Sulfate

DochEvi [55]

Answer:

Cyanide

Explanation:

<em>Molecular Structure of Each Answer</em>

A: CN-

B: NO3-

C: OH-

D: SO4 2-

As you can see, only A (Cyanide) is the only compound that does not contain oxygen, meaning it is NOT an oxyanion.

7 0

3 years ago

When a piece of metal A is placed in a solution containing ions of metal B, metal B plates out on the piece of A.(c) Which metal

Ostrovityanka [42]

Zinc is no longer the positive electrode because copper has a more positive (higher) value than zinc (anode). The anode value is reduced by the potential of the other electrode.

<h3>In a galvanic cell, is the anode positive or negative?</h3>

In a galvanic (voltaic) cell, the cathode is regarded as positive and the anode as negative. This seems reasonable given that the cathode is where electrons flow from the anode, which is where they originate.

<h3>What is a galvanic cell?</h3>

An electrochemical cell called a galvanic cell or voltaic cell, respectively named after the scientists Luigi Galvani and Alessandro Volta produces an electric current by spontaneous oxidation-reduction reactions. A typical device typically consists of two distinct metals that are submerged in separate beakers that each contains their own metal ions in solution and are either connected by a salt bridge or divided by a porous membrane.

Learn more about Galvanic cells here:-

brainly.com/question/13927063

#SPJ4

7 0

2 years ago

Write your own example of how the carbon cycle works.

bulgar [2K]

Answer:

Under the picture

Explanation:

7 0

3 years ago

Read 2 more answers

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .