According to the law of conservation of energy, which is always true

What is the chemical name of the compound K2SO4? Use the list of polyatomic ions and the periodic table to help you answer.

Ket [755]

potassium sulfate or A because K2 is potassium and SO4 is sulfate

6 0

2 years ago

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In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha

devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0

3 years ago

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Why there are different branches of chemistry

fredd [130]

The 5 branches are organic, inorganic, biochemistry, analytical, and physical. There are different branches because they each apply to different parts of chemistry. For example organic chemistry is the study of carbon compounds and inorganic chemistry is the study of non carbon compounds. You can't just have one type of chemistry because these are very different.

5 0

2 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

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

7 0

2 years ago

Which reaction is an example of an acid-base reaction? Which reaction is an example of an acid-base reaction?

oksian1 [2.3K]

Answer:

B) H2SO4 (aq) + Ca(OH)2 (aq) → CaSO4 (aq) + 2 H2O(l)

Explanation:

A is a reaction between a salt FeCl3 and a base KOH

C is a n acid decomposing on it's own to form two products

D is mercury, a metal reacting with oxygen. Two elements reacting. Neither are an acid or a base

E. is an acid reacting with a metal to liberate hydrogen. There is no base

3 0

2 years ago

According to the law of conservation of energy, which is always true​

According to the law of conservation of energy, which is always true