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2 years ago

470,809 mi express in an integer

Chemistry

2 answers:

navik [9.2K]2 years ago

8 0

Answer: The given number is already an integer.

Explanation:

Integer is defined as a whole number which is not a fraction. This number can be positive, negative or zero.

Decimal numbers are not considered as integers. They are rounded off to write an integer.

For Example: 4, -2 and 0, all are considered as integers.

We are given a number having value 470,809 miles. The number is already a whole positive number.

Thus, the given number is already an integer.

cupoosta [38]2 years ago

7 0

-470,809 m is in an integer

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Type the correct answer in the box. Express your answer to three significant figures.

satela [25.4K]

Answer:

The partial pressure of argon in the jar is 0.944 kilopascal.

Explanation:

Step 1: Data given

Volume of the jar of air = 25.0 L

Number of moles argon = 0.0104 moles

Temperature = 273 K

Step 2: Calculate the pressure of argon with the ideal gas law

p*V = nRT

p = (nRT)/V

⇒ with n = the number of moles of argon = 0.0104 moles

⇒ with R = the gas constant = 0.0821 L*atm/mol*K

⇒ with T = the temperature = 273 K

⇒ with V = the volume of the jar = 25.0 L

p = (0.0104 * 0.0821 * 273)/25.0

p = 0.00932 atm

1 atm =101.3 kPa

0.00932 atm = 101.3 * 0.00932 = 0.944 kPa

The partial pressure of argon in the jar is 0.944 kilopascal.

5 0

2 years ago

We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?

Anna11 [10]

M=D*V
D=620 g/cm³
V=75 cm³

m= 620 g/cm³ * 75 cm³=46500 g
m=46500g

8 0

2 years ago

A vessel contains Ar(g) at a high pressure. Which of the following statements best helps to explain why the measured pressure is

Luda [366]

Answer:

4. The combined volume of the Ar atoms is too large to be negligible compared with the total volume of the container.

Explanation:

Deviations from ideality are due to intermolecular forces and to the nonzero volume of the molecules themselves. At infinite volume, the volume of the molecules themselves is negligible compared with the infinite volume the gas occupies.

However, the volume occupied by the gas molecules must be taken into account. Each molecule does occupy a finite, although small, intrinsic volume.

The non-zero volume of the molecules implies that instead of moving in a given volume V they are limited to doing so in a smaller volume. Thus, the molecules will be closer to each other and repulsive forces will dominate, resulting in greater pressure than the one calculated with the ideal gas law, that means, without considering the volume occupied by the molecules.

5 0

2 years ago

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

Aleksandr-060686 [28]

To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s

5 0

2 years ago

Read 2 more answers

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30

FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) (1)

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) (2)

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

$10^{-1.30} = \frac{x^2}{0.09-x}$

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

pH = 0,0301

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

pH = 1,30

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

$10^{-6.70} = \frac{x^2}{0.09-x}$

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

pH = 2,87

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

pH = 6,70

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

$5.01x10^{-8} = \frac{x^2}{0.09-x}$

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

pH = 10,7

I hope it helps!

6 0

2 years ago