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kenny6666 [7]
3 years ago
5

Consider the energy diagram below.

Chemistry
2 answers:
Kazeer [188]3 years ago
4 0

Answer:

A) The catalyzed reaction passes through C.

Explanation:

nordsb [41]3 years ago
3 0

Answer:

The catalyzed reaction passes through C.

Explanation:

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professor190 [17]
D
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those are the correct answers
8 0
3 years ago
Next to each formula, write the number of atoms of each element found in one unit of the compound
satela [25.4K]
To answer this question I would have to know the elements in the compound
7 0
4 years ago
Arrange the forms of electromagnetic radiation in order of decreasing energy (from highest energy to lowest energy). You are cur
solmaris [256]

Answer:

gamma rays > X-rays > ultraviolet radiation > visible light > infrared > radio waves.

Explanation:

Electromagnetic waves are those waves that require no material medium for propagation. They can travel through space and they all move at the speed of light.

Electromagnetic waves are composed of both electric and magnetic fields which are mutually at right angles to each other.

The order of decreasing energy of electromagnetic waves is;

gamma rays > X-rays > ultraviolet radiation > visible light > infrared > radio waves.

3 0
3 years ago
Please help, super confused and urgent!
Lilit [14]

Empirical Formulae for;

Compound 1- K5 Mn5 O16

Compound 2- Na2 Cr2 O7

Compound 3- C3 H4 O4

Compound 4- C3 H3 O1

Explanation:

Step 1; as all the element percentages are given in percent assume the total mass of the compound is 100g and take each percentage as grams i.e., 27.5% of K as 27.5 g of K and so on.

Step 2; convert the mass of each element into their mole values by dividing available mass by molar masses.

Molar masses of required elements are as follows; K=39, Mn=55, O=16 C=12, H=1, Na=23, Cr=52.

Step 3; Divide all the values by the smallest mole value. I.e. for compound 1 after dividing the masses by molar masses we get 0.705, 0.681, and 2.187 for elements K, Mn, O respectively. Divide all three values with the least value which is 0.681 and write these values down.

Step 4; Convert all the numbers available into whole numbers by multiplying with suitable values. i.e. 3 if values are 0.33, 2 if values are 0.5 etc.  

Step 5; Assign these values to corresponding elements and you will get the above empirical formula.

4 0
4 years ago
The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide w
denis23 [38]

Answer:

a) 0,5 mol O₂; 1 mol NO₂

b)

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) NO = 30g

O₂ = 16g

NO₂ = 46g

d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO ×\frac{1molO_{2}}{2molNO} = <em>0,5 mol of O₂</em>

And produced:

1 mol NO ×\frac{2molNO_{2}}{2molNO} = <em>1 mol of NO₂</em>

b) By ideal gas law:

V = nRT/P

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO×\frac{30 g}{1molNO} = <em>30g</em>

0,5 mol O₂×\frac{32 g}{1molO_{2}} = <em>16g</em>

1 mol NO₂×\frac{46 g}{1molNO_{2}} = <em>46g</em>

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl×\frac{36,46 g}{1molHCl} = <em>7,41 g of HCl</em>

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: <em>107,7 g/mol</em>

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃×\frac{1mol}{122,55 gKClO_{3}} = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃×\frac{3 mol O_{2}}{2 molKClO_{3}} = <em>0,0312 moles of O₂</em>

In grams:

0,0312 moles of O₂ ×\frac{32g}{1 molO_{2}} = <em>0,999g of O₂</em>

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; <em>V = 0,699L ≡ 699mL</em>

At 29 °C (302,15K) and 732 torr (0,963 atm)

<em>V = 0,803L ≡ 803mL</em>

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂×\frac{2 mol KClO_{3}}{3 molO_{2}} = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃×\frac{122,55 g}{1molKClO_{3}} = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = <em>51%</em>

g. Combined gas law says that:

\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) <em>V₂ = 32,6 L</em>

<em></em>

I hope it helps!

4 0
3 years ago
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