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3 years ago

The most common physical state of the elements is

Chemistry

2 answers:

finlep [7]3 years ago

8 0

Answer: Solids, gases, or liquids

Explanation:

Matter can undergo physical and chemical changes called Phase Changes. Substances are classified based on physical states, often referred to as the States of Matter. At room temperature and pressure the common Physical States of elements are solids, gases or liquids.

Most are solids only 11 are gases and 2 are liquids

Ratling [72]3 years ago

3 0

Answer:

Solids.

Explanation: Most elements are solids, only 11 are gases and 2 are liquids.

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Vilka [71]

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3 years ago

7. A student has a 25 g sample of gas in a container at STP and adds another 7

BigorU [14]

Answer:

New temperature t2 = [1.28T− 273.15]° C

Explanation:

Given:

Volume v1 = 25 gram

New volume v2 = 25 + 7 = 32 gram

Constant pressure = p

Temperature t1 = T

Find:

New temperature t2

Computation:

Pv1/t1 = Pv2/t2

25 / T = 32 / t2

t2 = 1.28T

New temperature t2 = [1.28T− 273.15]° C

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3 years ago

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

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3 years ago

For the reaction ? Fe+? H2o ⇀↽? Fe3o4+? H2 , a maximum of how many grams of fe3o4 could be formed from 354 g of fe and 839 g of

Evgesh-ka [11]

The given reaction is:

3Fe + 4H2O → Fe3O4 + 4H2

Given:

Mass of Fe = 354 g

Mass of H2O = 839 g

Calculation:

Step 1 : Find the limiting reagent

Molar mass of Fe = 56 g/mol

Molar mass of H2O = 18 g/mol

# moles of Fe = mass of Fe/molar mass Fe = 354/56 = 6.321 moles

# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles

Since moles of Fe is less than H2O; Fe is the limiting reagent.

Step 2: Calculate moles of Fe3O4 formed

As per reaction stoichiometry:

3 moles of Fe form 1 mole of Fe3O4

Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4

Step 4: calculate the mass of Fe3O4 formed

Molar mass of Fe3O4 = 232 g/mol

# moles = 2.107 moles

Mass of Fe3O4 = moles * molar mass

= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)

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3 years ago

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The amplitude of a wave is the distance between a point on one wave and the identical point on the next wave T or F?

MaRussiya [10]

The amplitude of a wave is the distance between a point on one wave and the identical point on the next wave. The period and wavelength of a wave are inversely proportional.

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3 years ago