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TEA [102]
3 years ago
6

Which describes funguslike protists? They are all autotrophs. They are all able to move at some point. Their pseudopods allow th

em to reproduce. Their cells do not have cell walls.
Chemistry
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

Their pseudopods allow them to reproduce.

You might be interested in
State the oxidation number of S in <br><img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3
Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

  • The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
  • The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

<u><em>oxidation numbers of S= 4</em></u>

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

brainly.com/question/8990767

brainly.com/question/6498977

5 0
2 years ago
Read 2 more answers
A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of th
nekit [7.7K]
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


3 0
3 years ago
Read 2 more answers
Physical Science Structures and Properties of Matter Portfolio
daser333 [38]

Answer:

true

Explanation:

8 0
3 years ago
23) A common reaction that occurs in cells is shown here. In the presence of oxygen, a glucose molecule is combusted to form car
maw [93]

Answer:

C

Explanation:

add them together and multiply by 2

5 0
2 years ago
Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents
Bogdan [553]

The question is incomplete. There's missing the image, which is shown below.

Answer:

Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L

Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

P*12 = 8*RT

RT/P = 12/8 = 1.5

Thus, for O₂:

PV= nRT

V = n*(RT/P)

V = 4*1.5 = 6 L

For the mixture:

V = 12*1.5 = 18 L

For H₂O:

V = 8*1.5 = 12 L

The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L

6 0
3 years ago
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