All categories

3 years ago

Formula of diamond silver chloride?

1 answer:

3 0

AgCl is the chemical formula for silver chloride. There is no such thing as diamond silver chloride, unless you mean crystalline, but in that case, it still would be AgCl.

You might be interested in

3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.

igor_vitrenko [27]

Volume percent = Volume of solute
----------------------------------
Volume of the solution

   2 Volume of the solute
------- =   ------------------------------
100 250

Volume of the solute = 2 x 250
------------
100

= 5 mL.

Hope this helps!

4 0

3 years ago

A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra

Rasek [7]

The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

First-order reactions are very commonplace. we have already encountered examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁ = 0. 543 /s

T₁ = 25°C

Activation energy Eₐ = 75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁= Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.

Learn more about activation energy here:- brainly.com/question/26724488

#SPJ4

4 0

1 year ago

A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the

jenyasd209 [6]

Answer:

The volume of water in water bath is 1,011 Liters.

Explanation:

Length of the water bath, L = 1.85 m

Width of the water bath, W= 0.810 m

Height of the water bath ,H= 0.740 m

Height of the water in water bath, h= 0.740 m - 2.57 inches

1 m = 39.37 inch

$= 0.740 m - \frac{2.57}{39.37} m = 0.6747 m$

Volume of the water in bath = L × W × h

$=1.85 m\times 0.810 m\times 0.6747 m=1.011 m^3$

$1 m^3=1000 L$

$1.011 m^3=1.011\times 1000 L=1,011 L$

The volume of water in water bath is 1,011 Liters.

6 0

3 years ago

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac

Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0

3 years ago

A student preforms a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the m

tatuchka [14]

Answer:

Mass of water = 73.08 g

Explanation:

Given data:

Mass of hydrogen = 35 g

Mass of oxygen = 65 g

Mass of water = ?

Solution:

First of all we will write the balanced chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen = mass/ molar mass

Number of moles of hydrogen = 35 g/ 2 g/mol

Number of moles of hydrogen = 17.5 mol

Number of moles of oxygen = 65 g / 32 g/mol

Number of moles of oxygen = 2.03 moles

Now we compare the moles of water with moles hydrogen and oxygen.

H₂ : H₂O

2 : 2

17.5 : 17.5

O₂ : H₂O

1 : 2

2.03 : 2× 2.03 =4.06 mol

Number of moles of water produced by oxygen are less so oxygen is limitting reactant.

Mass of water:

Mass of water = number of moles × molar mass

Mass of water = 4.06 mol × 18 g/mol

Mass of water = 73.08 g

8 0

3 years ago