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1 year ago

Draw atomic depictions similar to those in Problem 2.43 for: (b) ⁷⁹₃₄Se

Chemistry

1 answer:

Murrr4er [49]1 year ago

5 0

The mass number of the element is 79 , atomic number is 34 and number of neutron is 45

<h3>What is Atomic Number ?</h3>

The number of proton which is used to separate one element from the other element is called Atomic Number. It is denotes as symbol Z.

What is Atomic Notation ?

The atomic number is the number of proton which is present in the nucleus of an atom. It is represented as $^{A}_{Z}X$ .

where

X is the element

Z is the atomic number of that element.

A is the mass number of that isotope.

Now,

A = Z + N

79 = 34 + N

79 - 34 = N

45 = N

Thus from the above conclusion we can say that mass number of the element is 79 , atomic number is 34 and number of neutron is 45

Learn more about the Atomic Number here: brainly.com/question/487717

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Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45

musickatia [10]

Answer:

For 1 antacid tablet (in ballon1) we get .0173 moles of CO2

for 2 tablets (in balloon 2) we get: 2*0,0173= 0.0346 moles of CO2

For 3 tablets (in balloon 3) we get 3* 0.0173 = 0.0519 moles of CO2

Explanation:

The complete question:

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

Balloon 1 had 1 antacid tab

Baloon 2 had 2

Balloon 3 had 3

Step 1: Data given

1.45 g of NaHCO3 is present in an antacid tablet

Molar mass of NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

NaHCO3 + H2O → NaOH + H2O + CO2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

1.45g / 84.0 g/mol = .0173 moles

Step 4: Calculate moles CO2

For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

so for 1 antacid tablet we get .0173 moles of CO2

for 2 tablets we get: 2*0,0173 = 0.0346 moles of CO2

For 3 tablets we get 3* 0.0173 = 0.0519 moles of CO2

5 0

3 years ago

Write a balanced equation for rusting. (Assume rust becomes Fe+2)

Stells [14]

Answer:

4Fe + 3O2 + 6H2O → 4Fe(OH)3

Explanation:

The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.

8 0

2 years ago

Can anyone here help me with chemistry

Kay [80]

The answer is 20g N2

5 0

2 years ago

Which of the conservationists discussed earlier in this unit is

Arte-miy333 [17]

The correct answer is A. The conservationist credited with founding the Wilderness Society is Bob Marshall.

Bob Marshall was an American forester, botanist, traveler, climber, writer and social activist who advocated the preservation of unspoiled wildlife.

He actively contributed to the development and adoption of environmental regulations prohibiting the construction of roads in vast federal specially protected natural areas. In 1935 he became one of the founders of the public environmental organization The Wilderness Society and its main sponsor.

Today Bob Marshall is considered one of the main initiators of the wildlife movement in the United States. Several geographic features are named after him, including the Bob Marshall Wilderness Nature Reserve in Montana and the Marshall Mountain on the Adirondack Ridge.

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6 0

2 years ago

Read 2 more answers

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago