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2 years ago

Write the atomic symbol for the isotope of bismuth with 125 neutrons.

Chemistry

1 answer:

vlada-n [284]2 years ago

5 0

First, isotopes are the atoms of a single element whose nuclei have a different number of neutrons, and therefore, differ in mass numbers. You should know that atoms are formed by a nucleus that has a small size and is made up of protons and neutrons. The nucleus is surrounded by a cloud of electrons, which are found in a region of the atom called the cortex.

The mass number, represented as A, is the sum of the number of protons and neutrons in the nucleus. On the other hand, the atomic number (Z) is the number of protons that exist in the nucleus.

The isotopes of an element X are represented as follows,

(see first attached picture)

It should be noted that the number of neutrons of a chemical element can be calculated as the difference A-Z.

The atomic and mass numbers of bismuth with 125 neutrons are:

Z = 83

A = 83 + 125 = 208

Thus, the atomic symbol of the bismuth isotope with 125 neutrons is:

(see second attached picture)

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In the following reaction, C6H6)? 2C6H6 + 15O2 12CO2 + 6H2O,

shepuryov [24]

The answer is 27.92g . working is shown in the picture above.

3 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago

26. Balance the following equations: Ca(s) + H3PO4(aq)Ca3(PO4)2(s) + H2(g)

yKpoI14uk [10]

Hey there!

Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance PO₄.

1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance H.

6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Balance Ca.

1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Our final balanced equation:

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Hope this helps!

3 0

3 years ago

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sergeinik [125]

Answer:

hulaan mo nlng

Explanation:

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2 years ago