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2 years ago

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A gas at 155 kilopascals and 25 °c has an initial volume of 1 liter the pressure of the gas increases to 605 kilopascals as the

temperature is raises ro 125°c what is the new volume?

1 answer:

KonstantinChe [14]2 years ago

3 0

i need to know the choices

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The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.

mafiozo [28]

As,
CuCO₃   ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² =  1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

x = 1.45 × 10⁻³ g/L

3 0

2 years ago

Water has a slight negative charge around its oxygen atom, and a slight positive charge around its hydrogen atoms Which type of

allochka39001 [22]

C.
Water is polar because one side of the molecule is positive and the other is negative.

3 0

3 years ago

Read 2 more answers

The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas

Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm$

Density is equal to mass per unit volume.

$d=\dfrac{m}{V}$

m is mass of the cylinder

V is the volume of the cylinder

$V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3$

So,

$m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g$

So, the mass of the cylinder is 1.51 grams.

5 0

3 years ago

1. 90.0 mL of distilled water is added to a 10.0mL sample of 0.150mol/L sodium

aniked [119]

Answer: dilute

Explanation:

A concentrated solution which is used to prepare solutions of lower concentrations by diluting it with addition of water.

A dilute solution is one which contains lower concentration.

Using Molarity equation:

$M_1$ =concentration of stock solution = 0.150 mol/L

$V_1$ = volume of stock solution = 10.0 ml

$M_2$ = concentration of dilute solution = ?

$V_2$ = volume of dilute solution = (10.0+90.0) ml = 100.0 ml

$0.150\times 10.0=M_2\times 100.0$

$M_2=0.015mol/L$

As the concentration is less than the original concentration, the solution is termed as dilute.

7 0

2 years ago

Balance the following reaction: AL2O3 + HCI ⇒ ALCI3 + H2O

Anastaziya [24]

To balance the following chemical equation, make a tally or a count of each of the atoms on both sides of the reaction, and make sure that those atoms are equal on both the reactant and product side.

AL2O3 + HCl => ALCl3 + H2O

Left side. Right side
AL = 2. AL = 1
O = 3 Cl = 3
H = 1. H = 2
Cl = 1. O = 1

First balance the metal atoms, aluminum, then hydrogen and then oxygen.

Balanced equation :
AL2O3 + 6HCl => 2ALCl3 + 3H2O.

Left side. Right side
AL = 2 AL = 2
O = 3 Cl = 6
H = 6 H = 6
Cl = 6 O = 3.

4 0

2 years ago