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3 years ago

When a chemical substance has either lost an electron or gained an oxygen?

1 answer:

7 0

if atoms want to be constant, they should gain or loose electrons from their valence orbits.
for doing this, metals will loose electrons and non-metals will gain electrons from metals. this happens in ionic compounds.

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Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio

soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂ → CuCO₃ + 2 NaCl

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:

Na₂CO₃: 1 mole
CuCl₂: 1 mole
CuCO₃: 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

Na₂CO₃: 129 g/mole
CuCl₂: 134.45 g/mole
CuCO₃: 123.55 g/mole
NaCl: 58.45 g/mole

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

Na₂CO₃: 1 mole ×129 g/mole= 129 grams
CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

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6 0

10 months ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

2 years ago

Convert 15.6 g Na2Cr2O7 to moles of Na2Cr2O7.

Leviafan [203]

Answer:

261.96754

Explanation:

3 0

3 years ago

A person is starting to run to get in shape. How would this runner’s time to complete a 10K race compare with that of a typical

Alexus [3.1K]

someone who has trained and is used to running will complete the 10k race faster than someone who just started running and not in as good of shape. They are both running a 10k race so the distance will be the same.

your answer should be

B) the time would be longer

7 0

3 years ago

You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t

Nataly [62]

First, we should get moles acetic acid = molarity * volume

=0.3 M * 0.5 L

= 0.15 mol

then, we should get moles acetate = molarity * volume

= 0.2 M * 0.5L

= 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity * volume

= 1 M * 0.02L

= 0.02 mol

when the reaction equation is:

CH3COOH + OH- → CH3COO- + H2O

moles acetic acid after adding OH- = (0.15-0.02)

= 0.13M

moles acetate after adding OH- = (0.1 + 0.02)

= 0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

= 0.13mol / 0.52L

= 0.25 M

and [acetate ) = 0.12 mol / 0.52L

= 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka

= -㏒ 1.8 x 10^-5

= 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

= 4.7 + ㏒(0.23/0.25)

= 4.66

6 0

2 years ago