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Jobisdone [24]
3 years ago
11

Help plz will mark brainliest xD

Chemistry
2 answers:
Paladinen [302]3 years ago
8 0

Answer:

Ions are atoms that have either gain or lose electrons from the neutral atom.

A loss of electrons results in a positive ion, called a cation, a gain of electron results in a negative ion called an anion.

Ionic compounds are formed by the combination of a cation and anion, or metal + nonmetal.

When naming ionic compounds, name the cation first, and the anion second, changing the ending to ide.

Explanation:

Hope it helped!

atroni [7]3 years ago
8 0

Answer:

• [ <u>lost</u> ] or [ <u>gained</u> ]

• [ <u>cation</u> ], [ <u>anion</u> ]

• [ <u>metal</u> ] + [ non-metal ]

• [ <u>cation</u> ] and the [ <u>anion</u> ] to [ <u>compound</u><u> </u><u>with</u><u> </u><u>suffix</u><u> </u><u>"</u><u>ide</u><u>"</u><u> </u><u>or</u><u> </u><u>"</u><u>ode</u><u>"</u> ]

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compact fluorescent light

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3 years ago
Need little help pls
Julli [10]

Answer:

0.354 molal

Explanation:

The molarity (M) or the concentration of a solution is defined as the number of moles of a compound in the solution per liter of solution (mol/L), whereas molality, is defined as the number of moles of a compound in the solution per kg of the compound (mol/kg).

Given that the density of the solution is 1.202 g/mL, which is equivalent to 1.202 kg/L. Since the prefix mili- denotes a factor of one thousandth ( 10^{-3} ) and kilo- denotes a factor of one thousand ( {10}^3 ),

1.202 \ \frac{g}{mL} \ = \ 1.202 \ \frac{g}{(10^{-3}) L} \ = \ 1.202 \ \frac{{10}^3g}{L} \ = \ 1.202 \ \frac{kg}{L}.

To calculate the corresponding molality of the solution, the formula

Molality \ (mol/kg) \ = \ \frac{Molarity \ (mol/L)}{Density \ (kg/L)} is used.

Therefore,

Molality \ = \ \frac{0.426 \ mol/L}{1.202 \ kg/L} \ = \ 0.354 \ mol/kg \ \ (3 \ s.f.)

4 0
3 years ago
MULTIPLE CHOICE QUESTION
Naddika [18.5K]

Answer:

rust

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4 0
2 years ago
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Answer:

I'm pretty sure that's right.

Explanation:

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8 0
3 years ago
A gas mixture at room temperature contains 10.0 mol CO and 12.5 mol O2. (a) Compute the mole fraction of CO in the mixture. (b)
zlopas [31]

Answer:

a. the mole fraction of CO in the mixture of CO and O2.

mole fraction = moles of CO/ Total moles of the mixture

Mole fraction of CO = 10/(10+12.5)=0.444

b. Reaction - CO(g)+½O2(g)→CO2(g)

Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2

So given,

At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.

3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2

This means that unused mols are : 7mols of CO and 11mols of O2

Total product mixture = 3 + 7 + 11 = 21mols

mole fraction of CO = 7/21 = 0.33

7 0
3 years ago
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