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Charra [1.4K]
3 years ago
13

If a cell wants to tell another cell what to do, it will send the message through

Chemistry
1 answer:
horsena [70]3 years ago
6 0

Answer:

it will send the message through the axon

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What major product is formed when heptyne is treated with Br2 in CCl4?
ANTONII [103]

Answer:A 1,2-dibromoheptene

Explanation:

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3 years ago
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How many grams of potassium chloride are required to make 250 ml of a 0.75 m kcl solution?
DedPeter [7]

Answer:

The solution would need 13.9 g of KCl

Explanation:

0.75 m, means molal concentration

0.75 moles in 1 kg of solvent.

Let's think as an aqueous solution.

250 mL = 250 g, cause water density (1g/mL)

1000 g have 0.75 moles of solute

250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl

Let's convert that moles in mass (mol . molar mass)

0.1875 m . 74.55 g/m = 13.9 g

7 0
2 years ago
Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?
Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.  

The number of silver atoms  13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.  

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0
3 years ago
What is the name of the pores, or holes, on the outer layer of the sponge that take water in?
ivanzaharov [21]
The tiny holes in a sponges outer layer that take in water are called ostia, I believe. Hope it helps
7 0
3 years ago
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