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tankabanditka [31]
2 years ago
13

Which of these organisms reproduce through binary fission? (Select all that apply.)

Chemistry
1 answer:
tensa zangetsu [6.8K]2 years ago
7 0

Amoeba are those Organisms which reproduce through binary fission.

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5 Na molecules and 5 Cl molecules
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Why are the densities of corn syrup and gasoline expressed as a range of values
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<span> because gasoline changes volume as a function of temperature or because there are different grades of gasoline or because the values are given in different units of measure .</span>
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3 years ago
Could someone help me with this?
Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

  • Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

  • x is the oxidation number

#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

7 0
2 years ago
H2S what species are present at 10-6 mol/L or greater when dissolved in water
Anastaziya [24]

Answer:

the answer has been given below have a good day

Explanation:

mark as brainiest

5 0
2 years ago
If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density
adelina 88 [10]

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are 6.023 \times 10^{26} atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = \frac{38}{6.023 \times 10^{26}}    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = \frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}

            = 37.06 \times 10^{-26}

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = a^{3}

                   = (0.503 \times 10^{-9})^{3}

                   = 0.127 \times 10^{-27} m^{3}

Formula to calculate density of diamond cell is as follows.

               Density = \frac{mass}{volume}

                             = \frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}

                            = 2918.1 g/m^{3}

or,                         = 0.0029 g/cc       (as 1 m^{3} = 10^{6} cm^{3})

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

4 0
3 years ago
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