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Nata [24]
2 years ago
7

What is a chemical reaction?

Chemistry
2 answers:
Mila [183]2 years ago
8 0

Answer:

A chemical reaction is defined as per encyclopedia Britannia as a "a process in which one or more substances, are converted to one or more different substances, the products."

So A is correct, B is correct, C is correct.

So that means we choose D as they're all correct.

Mkey [24]2 years ago
5 0

Answer:

D.

Explanation:

this is because in a chemical reaction all 3 of those work.

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What is the answer to 1 and 2?
ankoles [38]
You got the questions correct
4 0
3 years ago
A beam of light has a wavelength of 24.0 μm ( μm = 10 -6m). What is its frequency?
vfiekz [6]

Answer:

Frequency = 1.25 ×10¹³ Hz

Explanation:

Given data:

Wavelength of light = 24.0  μm (2.4 ×10⁻⁵ m)

Frequency = ?

Solution:

Formula:

Speed of light = wavelength × frequency

Speed of light /wavelength = frequency

Frequency = 3×10⁸ m /s /2.4 ×10⁻⁵m

Frequency = 1.25 ×10¹³ s⁻¹

s⁻¹ = Hz

Frequency = 1.25 ×10¹³ Hz

6 0
2 years ago
C= ( F - 32) / 1.8<br> 86 F = ?
egoroff_w [7]

<h3>⇒c =  \frac{f - 32}{1.8}</h3><h3>⇒c =  \frac{86 - 32}{1.8}</h3><h3>⇒c =  \frac{54}{1.8}</h3><h3>⇒c = 30</h3>

<em>Hence</em><em>,</em><em> </em><em>84</em><em>°</em><em>F</em><em> </em><em>is</em><em> </em><em>30</em><em>°</em><em>C</em><em> </em><em>in</em><em> </em><em>Celsius</em><em>.</em><em>.</em>

7 0
2 years ago
14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d
zepelin [54]

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock solution.

M_2\text{ and }V_2 are the molarity and volume of diluted solution.

From data (A) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

5 0
2 years ago
During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that
bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

4 0
2 years ago
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