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user100 [1]
3 years ago
11

Determine the number of protons and electrons for La3+

Chemistry
1 answer:
Yanka [14]3 years ago
7 0

Answer:

Ag+ = 47 electrons - 1 electron = 46 electrons. Finally: 47 protons , 61 neutrons and 46 electrons.

Explanation:

You might be interested in
6.50 g of a certain Compound x, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1
gayaneshka [121]

Answer:

C_{10}H_8

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to find the molecular formula of the given compound by firstly calculating both moles and grams of carbon in carbon dioxide and hydrogen in water, as the only sources of these elements derived from the compound x due to its combustion:

n_C=22.35gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.51molC\\\\m_C=0.51molC*\frac{12.01gC}{1molC}   =6.10gC

n_H=3.66gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.41molH\\\\m_H=0.41molH*\frac{1.01gH}{1molH}=0.41g

Now, since the addition of carbon and hydrogen is about 6.50 grams, we infer the compound has no oxygen, that is why we now set the mole ratios in the empirical formula for both C and H as shown below:

C:\frac{0.51mol}{0.41mol}= 1.24\\\\H:\frac{0.51mol}{0.51mol}= 1\\\\C_{1.24}H

Yet it cannot be decimal, that is why we multiply by 4 to get the correct whole-numbered empirical formula:

C_5H_4

Whose molar mass is 64.09 g/mol, which makes the ratio of molar masses:

\frac{128.g/mol}{64.09g/mol} =2

Therefore, the molecular formula is twice the empirical one:

C_{10}H_8

Regards!

8 0
3 years ago
When gases are compressed, the particles are forced closer together.
Elanso [62]
Gases can be compressed easily because there is a large amount of space between the individual molecules, which are very active and move around at high speed. When gases are compressed, the particles are forced much closer together, allowing a huge amount of particles to fit a small space.
5 0
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The density of helium in a balloon is 1.18 g/L. If a
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There is 3.58 He in the balloon.
3 0
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Why is carbon dioxide used in fire extinguisher?
rjkz [21]
As Fire needs oxygen to burn, Carbon Dioxide Covers the fire like an blanket and cuts the contact of fire to the oxygen present in the air, So Fire stops burning.That is why Carbon Dioxide is Considere as the best Fire Extinguisher.
3 0
3 years ago
Read 2 more answers
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
timama [110]

Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

8 0
3 years ago
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