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ICE Princess25 [194]
2 years ago
5

Sponges reproduce using: A mitosis B budding C sexual reproduction

Chemistry
2 answers:
kupik [55]2 years ago
4 0
B, budding, good luck!
vlada-n [284]2 years ago
3 0

Answer:

It is B. Budding

Explanation:

Hope this helped have an amazing day!

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1#
Black_prince [1.1K]

Answer:

1= Magnesium

2 = Option 3 =  1s² 2s² 2p⁶

Explanation:

An electrically neutral atom consist of equal number of protons and electrons.

The answer for 1st q is magnesium because the electronic configuration showed twelve number of electrons. The atomic number of magnesium is twelve that's why this configuration is of Mg.

Mg₁₂ =  1s² 2s² 2p⁶ 3s²

The second answer is option three because has atomic number ten and third electronic configuration have ten electrons.

Ne₁₀ = 1s² 2s² 2p⁶

It is stable electronic configuration. Neon is inert because of this electronic configuration. The outer most shell is completely filled.

8 0
3 years ago
______ have no definite shape, but definite volume
NemiM [27]

Answer:

liquids

Explaination:

3 0
3 years ago
The endocrine system is made of which components?
AysviL [449]
Adrenal Glands
<span>Pancreas</span>
<span>Thyroid and Parathyroid Glands</span>
3 0
3 years ago
For which one of the following molecules is the indicated type of hybridization not appropriate for
Aleksandr [31]

A. BeCl2 sp2

Also when you get the chance, could you mark me brainliest?

7 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
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