What happens to the atomic radius when an electron is gained?

Chemistry

1 answer:

LenaWriter [7]2 years ago

6 0

A. The negative ionic radius is larger than the neutral atomic radius.

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What is per cent by mass of Na2CO3 in a 1.0 molal aqueous solution ?

aivan3 [116]

Hello!

We have the following data:

MM (Molar mass of Na2CO3)
Na = 2*23 = 46 u
C = 1*12 = 12 u
O = 3*16 = 48 u
----------------------
MM (Molar mass of Na2CO3) = 46+12+48 = 106 g/mol
n (mol number) = 1 mol
m1 (mass of the solute) = ?

$n = \dfrac{m_1}{MM}$

$1\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}= \dfrac{m}{106\:g/mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$m_1 = 1*106$

$\boxed{m_1 = 106\:g}\Longleftarrow(solute\:mass)$

So if we have the dissolution in a liter (m2 - mass of solvent), that is 1000g , then the total mass of the solution (m) will be:

$m\:(solution\:mass) = m_1\:(solute\:mass) + m_2\:(solvent\:mass)$

$m = 106\:g+1000\:g$

$\boxed{m = 1106\:g}\Longleftarrow(solution\:mass)$

Now, let's find the percentage in mass (% m / m), let's see:

$\%(m/m) = \dfrac{m_1}{m}*100$

$\%(m/m) = \dfrac{106}{1106}*100$

$\%(m/m) = \dfrac{10600}{1106}$

$\%(m/m) = 9.58408...\to\:\boxed{\boxed{\%(m/m) \approx 9.6}}\end{array}}\qquad\checkmark$

Answer: B) 9.6

I Hope this helps, greetings ... DexteR!

5 0

3 years ago

How many types of endoplasmic recticulum are there in a cell?

Yuki888 [10]

Answer:

Two

Explanation:

6 0

3 years ago

Read 2 more answers

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below. 2HgO Rig

BigorU [14]

Taking into account the reaction stoichiometry, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.

In first place, the balanced reaction is:

2 HgO → 2 Hg + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

HgO: 2 moles
Hg: 2 moles
O₂: 1 moles

The molar mass of the compounds is:

HgO: 216.59 g/mole
Hg: 200.59 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

HgO: 2 moles× 216.59 g/mole= 433.18 grams
Hg: 2 moles× 200.59 g/mole= 401.18 grams
O₂: 1 mole× 32 g/mole= 32 grams

Then the following rule of three can be applied: if by reaction stoichiometry 32 grams of O₂ are produced by 2 moles of HgO, 250 grams of O₂ are produced from how many moles of HgO?

$moles of HgO=\frac{250 grams of O_{2} x2 moles of HgO}{32grams of O_{2}}$