All categories

3 years ago

How many grams of sodium are in 0.820 moles of na2so4?

2 answers:

5 0

the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na₂SO₄ is - 142 g/mol.
1 mol of Na₂SO₄ has a mass of 142 g.
In 1 mol of Na₂SO₄ the mass of Na is 23 g/mol x 2 = 46 g.

Mass of Na in 1 mol of Na₂SO₄ is - 46 g

mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g

Irina-Kira [14]3 years ago

3 0

Explanation:

Number of moles are defined as given mass divided by molar mass of the compound or substance.

Mathematically, Number of moles = $\frac{mass given}{molar mass}$

Since, it is given that number of moles is 0.820 and molar mass of $Na_{2}SO_{4}$ is 142 g/mol.

Therefore, calculate the mass as follows.

Number of moles = $\frac{mass given}{molar mass}$

0.820 moles = $\frac{mass given}{142 g/mol}$

mass given = 116.44 g

Thus, we can conclude that there are 116.44 grams of sodium in 0.820 moles of $Na_{2}SO_{4}$ .

You might be interested in

Please help... will give brainiest

Dvinal [7]

Answer:

The answer will be ：

In year 1869

Russian scientist named Dmitri Mendeleev

Chart called Mendeeleves periodic table

increasing atomic masses

physical and chemical properties.

Hope this Helps

Best of luck ！！

7 0

3 years ago

A magnet moved near a coil of wire can cause a(n)

Phantasy [73]

When a magnet moves near a coil of wire it can cause an A. electric current

7 0

4 years ago

Read 2 more answers

Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a

nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

= 75.3%

4 0

4 years ago

Pliz, answer my this all question

never [62]

18. a) The materials that are in contact. The two materials and the nature of their surfaces. ...
b) The force pushing the two surfaces together. Pushing the surfaces together causes the more of the asperities to come together and increases the surface area in contact with each other.

19. the quantity of motion of a moving body, measured as a product of its mass and velocity.

20. According to Newton's third law of motion, action force is equal to reaction but acts on two different bodies and in opposite directions. When a horse pushes the ground, the ground reacts and exerts a force on the horse in the forward direction. This force is able to overcome friction force of the cart and it moves.

21. The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

22. R12. Mass is more fundamental because it is an intrinsic property of an object. Weight varies with location depending upon the acceleration due to gravity eg. for a mass m = 10kg on Earth it`s weight is W = mg = 10 x 10 = 100N.

8 0

4 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

7 0

3 years ago