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gulaghasi [49]
3 years ago
7

How many grams of Sulfuric Acid are needed to produce 57.18 g of Lead (IV) Sulfate when being neutralized by a sufficient amount

of Lead (IV) Hydroxide? *
Chemistry
1 answer:
KiRa [710]3 years ago
7 0

Answer:

40.72g of sulfuric acid are needed

Explanation:

When sulfuric acid, H₂SO₄, is neutralized by lead (IV) hydroxide, Pb(OH)₄, Lead (IV) sulfate, Pb(SO₄)₂ and water as follows:

2 H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

To solve this question we must find the moles of 57.18g of Pb(SO₄)₂. As 2 moles of H₂SO₄ produce 1mol Pb(SO₄)₂ we can find the moles of H₂SO₄ and its mass as follows:

<em>Moles Pb(SO₄)₂ -Molar mass: 275.23 g/mol-</em>

57.18g * (1mol / 275.23g) = 0.2078 moles Pb(SO₄)₂

<em>Moles H₂SO₄:</em>

0.2078 moles Pb(SO₄)₂ * (2mol H₂SO₄ / 1mol Pb(SO₄)₂) = 0.4155 moles H₂SO₄

<em>Mass H₂SO₄ -Molar mass: 98g/mol-</em>

0.4155 moles H₂SO₄ * (98g / mol) =

<h3>40.72g of sulfuric acid are needed</h3>
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A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70
kakasveta [241]

Explanation:

The given data is as follows.

  P_{atm} = 98.70 kPa = 98700 Pa,  

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

      height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = 13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}

                = 13534 kg/m^{3}

The relation between pressure and atmospheric pressure is as follows.

             P = P_{atm} + \rho gh

Putting the given values into the above formula as follows.

            P = P_{atm} + \rho gh

               = 98700 Pa + 13534 \times 9.81 \times 0.03 m

               = 102683.05 Pa

               = 102.68 kPa

thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.

8 0
3 years ago
What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = 5\times 196.96 g

4.99 moles of gold contain = 984.8 g

Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

3 0
3 years ago
A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

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After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

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Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

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Answer:

is an example of precipitation from a solution.

Explanation:

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Answer:

D. All of the Above

Explanation:

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