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3 years ago

How many questions would you need to separate ten arthropods into different branches

Chemistry

1 answer:

Vikki [24]3 years ago

6 0

Answer:

10.

Explanation:

We need 10 question to separate ten arthropods into different branches because each arthropod belongs to separate branch and each branch needs explanation. Phylum Arthropoda is considered as the largest phylum in the animal world, and insects form the largest class within this phylum due to its highest number of individual or organisms. Each arthropods are different from one another so they needs more questions to separate them from each other.

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

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

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers

Briefly describe the solvation of sodium chloride to form aqueous solution.

charle [14.2K]

Answer:

idk

Explanation:

sorry

7 0

2 years ago

Read 2 more answers

A sample of kerosene has a mass of 36.4g. It’s volume is 45.6mL. What is the density of kerosene?

stira [4]

Answer:

<h2>Density = 0.8 g/cm³</h2>

Explanation:

The density of an object can be found using the formula

<h3> $Density(\rho) = \frac{mass}{volume}$ </h3>

From the question

mass of kerosene = 36.4 g

volume of kerosene = 45.6 mL

To find the density substitute the values into the above formula and solve

We have

<h3> $Density = \frac{36.4}{45.6}$ </h3>

= 0.7982

We have the final answer as

<h3>Density = 0.8 g/cm³</h3>

Hope this helps you

4 0

3 years ago

The transfer of heat energy that does not require matter of any type is _____.

S_A_V [24]

This type of heat transfer is radiation since we can experience the heat from an object without being in contact with it. Energy is being emitted by electromagnetic waves or in moving subatomic particles. Radiation is a consequence of thermal agitation of the molecules.

6 0

2 years ago

Read 2 more answers

Identify each element below, and give the symbols of the other elements in its group:

nikdorinn [45]

Answer:

In explanation

Explanation:

a. Selenium, atomic number 34

Argon is 18, adding the other electrons gives 34.

Other elements include: oxygen O , Sulphur S , Tellurium Te and Polonium Po

b. Hafnium, element 72

Xe is 54, adding the other electrons makes the total 72.

Other elements; Titanium Ti, Zirconium Zr, Rutherfordium Rf

c. Manganese, element 25

Other group members are Technetium Tc , Rhenium Re and Bohrium Bo

3 0

3 years ago