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wel
2 years ago
10

choose the true statement(s) about ions. to be marked correct, you’ll need to select all true statements, as there may be more t

han one correct answer.
Chemistry
1 answer:
mash [69]2 years ago
7 0

Cations have more protons than electrons, Anions result when atoms gain an electron.

<h3>What are electrons ?</h3>

The electron, a subatomic particle, has a negative one elementary charge electric charge. The electron, a member of the first generation of the lepton particle family, is usually regarded as an elementary particle because it has no known components or substructure.

<h3>How electrons are in an atom ?</h3>

A few straightforward rules can be used to calculate the quantity of protons, neutrons, and electrons present in an atom. The atomic number and the number of protons in the atom's nucleus are equal (Z). A neutral atom has the same number of protons and electrons as protons.

Around the nucleus, electrons are positioned in several shells. Only a fixed amount of electrons can fit in each subsequent shell. Filling begins with the innermost shell.

To know more about nucleus visit :

brainly.com/question/23366064

#SPJ4

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Find the maximum mass of MgO that can be made from 2.4 of Mg and 2.4 of O2
Yanka [14]
Need more information
3 0
3 years ago
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0
3 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
Which tool or tools would you use to measure and compare the mass of a cup of fresh water and a cup of salt water
Artemon [7]
You could use a scale to measure the mass as well as a cup to hold the water. If you were comparing the two, you should also probably use a graduated cylinder to get the same amount of each type of water.

Hope this helped ^_^
4 0
3 years ago
What is the change in boiling point for a 0.615m solution of Mgl2 in water?
ivanzaharov [21]

Answer :  The change in boiling point is, 0.94^oC

Explanation :

Formula used :

\Delta T_b=i\times K_f\times m

where,

\Delta T_b = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

K_f = boiling point constant for water = 0.51^oC/m

m = molality  = 0.615 m

Now put all the given values in this formula, we get

\Delta T_b=3\times (0.51^oC/m)\times 0.615m

\Delta T_b=0.94^oC

Therefore, the change in boiling point is, 0.94^oC

4 0
3 years ago
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