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Kitty [74]
2 years ago
13

Which is an example of a vascular flowering seeded plant? \

Chemistry
1 answer:
olga55 [171]2 years ago
6 0
One plant would be the angiosperm. A
vascular, seed-producing plants whose seeds are enclosed in a fruit.
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Atomic emission spectra are produced when the light emitted by an element passes through a prism. Which of the following stateme
xenn [34]
Option C: elements produce spectra with only few distinct lines.

The spectra are not continuos and are different for every element. This permits to identify elements.
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3 years ago
Which one of the following electron transitions does not correspond to an absorbtion of a quanta of light?
ziro4ka [17]
N = 4 to n = 3 is the right answer, so it' none of the above
8 0
3 years ago
What is the mass of 2.50 moles of NaCl
garik1379 [7]

Answer:

The mass of 2,50 moles of NaCl is 146, 25 g.

Explanation:

First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:

Weight NaCl= Weight Na + Weight Cl=  23 g+ 35,5 g= 58, 5 g/ mol

1 mol ------ 58, 5 g

2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>

8 0
3 years ago
Bradley was working in the lab. He accidentally poured a container of concentrated hydrochloric acid into his container of disti
nexus9112 [7]

Answer is: The solution has now become a good conductor of electricity.

Hydrochloric acid (HCl) dissociate on positive ions or cations of hydogen (H⁺) and negative ions or anions of chlorine (Cl⁻) accordinf to balanced chemical reaction:

HCl(aq) → H⁺(aq) + Cl⁻(aq).

When there are free cations and ions, water solution can conduct electricity.

4 0
2 years ago
Read 2 more answers
Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).
mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

             CuSO₄.5H₂O  →   CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

         Number of moles  = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

           Volume  = 250 x 10⁻³L

  Input the parameters and solve for number of moles;

        Number of moles  = 250 x 10⁻³  x  2.01 = 0.5mol

From the equation;

             1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O  

So  0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16)  = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

                                      = 0.5 x 249.6

                                     = 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0
3 years ago
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