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kogti [31]
3 years ago
14

Which of the following is an acid-base neutralization reaction? (1 point) Sn + 2HBr yields SnBr2 + H2 HCl + KOH yields KCl + H2O

2AlCl3 + 3Ca(OH)2 yields 2Al(OH)3 + 3CaCl2 2C2H6 +7O2 yields 4CO2 + 6H2O
Chemistry
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

Explanation:

Sn + 2HBr =  SnBr₂ + H₂

Here HBr is an acid but Sn is not a base . It is a metal . So it is not an acid - base reaction .

HCl + KOH =  KCl + H₂O

HCl is an acid and KOH is a base so it is an acid base reaction.

2AlCl₃ + 3Ca(OH)₂ =  2Al(OH)₃ + 3CaCl₂

It is an acid base reaction . It is so because aluminium hydroxide is a lewis acid and calcium hydroxide is a base . So it is an acid base reaction .

2C₂H₆ +7O₂ =  4CO₂ + 6H₂O

It is not an acid base reaction . It is actually an example of oxidation reaction in which ethane burns in oxygen to give carbon dioxide and water.

You might be interested in
How many atoms of aluminum are in one formula unit of aluminum chloride?
ICE Princess25 [194]

The formula for aluminum fluoride is AlF_{3}


That means that there is 1 Aluminum and 3 fluorde ions in 1 molecule. The 3 does not effect the Aluminum at all.


The Answer is A

7 0
3 years ago
A new penny has a mass of 2.49 grams and occupies 0.349 cm cubed. If pure copper has a density of 8.96 g/cm cubed, is the new pe
lys-0071 [83]

Answer: The new penny is not pure copper

Explanation:

Density \rho is defined as a relation between the mass m and the volume V:

\rho=\frac{m}{V}

Now, we are told the density of pure copper is:

\rho_{copper}=8.96 g/cm^{3}

And we are given the mass and volume of the new penny, with which we can calculate its density:

\rho_{penny}=\frac{m_{penny}}{V_{penny}}=\frac{2.49 g}{0.349 cm^{3}}

\rho_{penny}=7.13 g/cm^{3} As we can see the density of this penny is not equal to the density of pure copper, hence the new penny is not pure copper.

8 0
4 years ago
1500 millimeters to km
gizmo_the_mogwai [7]
0.0015 kilometers is for sure the answer!
4 0
3 years ago
What is the most mass of (NH4)2CO3 (H=1,C=12,N=14,O=16
Agata [3.3K]

Answer:

96.09 g/mol

Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

N = 14.01 g/mol

H = 1.008 g/mol

O = 16.00 g/mol

C = 12.01 g/mol

Since your compound is  

(

N

H

4

)

2

C

O

3

, you need to multiply the atomic weights by their subscripts. Therefore,

Step 2

N = 14.01 g/mol × 2 =

28.02 g/mol

H = 1.008 g/mol × (4×2) =

8.064 g/mol

 

O = 16.00 g/mol × 3 =

48.00 g/mol

C = 12.01 g/mol × 1 =

12.00 g/mol

To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

molar mass of

(

NH

4

)

2

CO

3

=

(28.02 + 8.064 + 48.00 + 12.01) g/mol

=

96.09 g/mol

this is a google search and a example i hope is helps to solve

6 0
3 years ago
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

6 0
3 years ago
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