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3 years ago

What is the atomic mass of the element based off from the charts ?

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu, also known as daltons, D). The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope.

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Which statements describe the Rutherford atomic model? Check all that apply.

MaRussiya [10]

Answer:

1 and 4

Explanation:

7 0

2 years ago

Read 2 more answers

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the

frosja888 [35]

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

5 0

3 years ago

5. Which are physical properties of a property of Aluminum metal?

4vir4ik [10]

Answer:

Aluminum is a shiny, silvery white colored metal that is light in weight and strong. Th density of aluminum is 2.7 g/mL, which means the metal will sink in water, but is still relatively light

Hope this helps (:

Mark me brainliest if its good for you :D take care.

8 0

2 years ago

A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa

finlep [7]

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

Moles KNO3 -Molar mass: 101.1032g/mol-

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

Change in heat:

q = m*S*ΔT

Where q is heat in J,

m is the mass of the solution: 10.6g + 251.0g = 261.6g

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

Molar heat of solution:

3830.87J/0.1048 moles KNO3 =

36554J/mol =

6 0

2 years ago

Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0g of potassium and

Leona [35]

Answer:

79.43kg

Explanation:

To adequately solve this problem, we should know the law of constant composition. This is a pointer to the fact that no matter the type of sample, the percentage compositions of potassium and iodine still remains the same. We can use these masses to get a formula and we would know the percentage compositions in whatever mass we are dealing with.

First of all, we add the masses in the first sample. 13 + 42.3 = 55.3

Hence, the percentage composition of the potassium is 13/55.3 * 100 = 23.51%

The percentage composition of the iodine is = 100 - 23.51 = 76.49%

Now, we need to get the formula of the compound. We can get this by dividing the percentage compositions with the atomic masses. The atomic mass of potassium and iodine is 39 and 127 respectively.

Potassium = 23.51/39 =0.603

Iodine = 76.49/127 = 0.602

We then divide by the smaller value to get the formula and this shoes our formula is KI

We can see they have a ratio of 1 to 1, meaning one atom of potassium to one atom of iodine. This further confirms the percentage compositions of 23.5 to 76.5

Now to get the mass of iodine yielded, let us say the mass is xkg

This means x/(x + 24.4) * 100 = 76.5

100x = 76.5( x + 24.4)

100x = 76.5x + 1866.6

100x - 76.5x = 1866.6

23.5x = 1866.6

x = 1866.6/23.5 = 79.43kg

4 0

2 years ago