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Nikolay [14]
2 years ago
11

Using correct significant figures, calculate the length of one side of a cube made of gold, if the mass of the cube is 4.05 g an

d the density of gold is 19.32 g/cm3
Chemistry
1 answer:
Nutka1998 [239]2 years ago
5 0

Answer:

Explanation:

density = 19.32

Mass = 4.05 g

V = ?

Formula

Density = mass / volume or

volume = mass / density.

Solution

volume = mass / volume

volume = 4.05 / 19.32

Volume = 2.10

The solid cube has a side length of ∛V which reads as the cube root of volume.

Side Length = ∛2.10 = 0.594 cm

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What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?
Natali5045456 [20]

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2  moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2  moles of P requires 5 moles of chlorine

0.109  moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

brainly.com/question/3637553

#SPJ4                      

5 0
1 year ago
A chemist requires 0.802 mol Na2CO3 for a reaction. How many grams does this correspond to?
Komok [63]

Answer:

Ok:

Explanation:

So grams = mols*MolarMass. Here, MolarMass (MM) = 105.99g which can be found using the periodic table. mols is given to be 0.802. We can then plug in to get that it corresponds to 85.0g.

7 0
2 years ago
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
Consider a general reaction
choli [55]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

7 0
3 years ago
How would granite be classified as
vredina [299]

Answer:

Granite is an intrusive igneous rock which means it is cooled slowly deep upper the Earth's crust. It is composed of 25% to 35% quartz and over 50% potassium- and sodium rich feldspars.

Explanation:

4 0
2 years ago
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