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lord [1]
4 years ago
8

Write down the new substance formed when ammonia react with hydrochloric acid​

Chemistry
1 answer:
kow [346]4 years ago
6 0

Answer: ammonium chloride

Explanation: The reaction equation among ammonia (NH3) and hydrochloric acid (HCl) is written as follows: NH3+HCl=NH4Cl. Ammonia is a susceptible base that reacts with hydrochloric acid, forming a compound referred to as ammonium chloride.

You might be interested in
How many moles of Chromium are in 1.410 * 10 ^ 23 atoms of Chromium?
leva [86]

Answer:

<h3>The answer is 0.23 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L}  \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{1.410 \times  {10}^{23} }{6.02 \times  {10}^{23} }  \\  = 0.234219...

We have the final answer as

<h3>0.23 moles</h3>

Hope this helps you

4 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
Read 2 more answers
How much excess reactant is left over when 17.0 g of potassium hydroxide (KOH) reacts with
dolphi86 [110]

Answer:

4.56 g of KOH

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KOH + Fe(NO₃)₂ —> Fe(OH)₂ + 2KNO₃

Next, we shall determine the masses of KOH and Fe(NO₃)₂ that reacted from the balanced equation. This is can be obtained as:

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

Mass of KOH from the balanced equation = 2 × 56 = 112 g

Molar mass of Fe(NO₃)₂ = 56 + 2[14 + (16×3)]

= 56 + 2[14 + 48)]

= 56 + 2[62]

= 56 + 124

= 180 g/mol

Mass of Fe(NO₃)₂ from the balanced equation = 1 × 180 = 180 g

SUMMARY:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂

Next, we shall determine the limiting reactant and the excess reactant. This can be obtained as follow:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂.

Therefore, 17 g of KOH will react with = (17 × 180)/112 = 27.32 g of Fe(NO₃)₂

From the calculations made above, we can see that it will take a higher mass (i.e 27.32 g) of Fe(NO₃)₂ than what was given (i.e 20 g) to react completely with 17 g of KOH.

Therefore, Fe(NO₃)₂ is the limiting reactant and KOH is the excess reactant.

Next, we shall determine the mass of the excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

112 g of KOH reacted with 180 g of Fe(NO₃)₂.

Therefore Xg of KOH will react with 20 g of Fe(NO₃)₂ i.e

Xg of KOH = (112 × 20)/180

Xg of KOH = 12.44 g

Thus, 12.44 g of KOH reacted.

Finally, we shall determine the leftover mass of the excess reactant.

The excess reactant is KOH. The leftover mass can be obtained as follow:

Mass of KOH given = 17 g

Mass of KOH that reacted = 12.44 g

Mass of KOH leftover =?

Mass of KOH leftover = (Mass of KOH given) – (Mass of KOH that reacted)

Mass of KOH leftover = 17 – 12.44

Mass of KOH leftover = 4.56 g

Thus, the excess reactant (i.e KOH) that is left over is 4.56 g

3 0
3 years ago
A student is doing a titration using potassium permanganate solution, KMnO4, to determine the amount of H2O2 in a sample. The ba
sergejj [24]

Answer:

The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water.

Explanation:

In a titration, the equivalent moles of titrant must be the same than analyte. If the student calculates an amount of moles of H₂O₂ that is larger than the actual value:

The student failed to wear goggles <em>FALSE. </em>The use of goggles doesn't change the results in the lab but you must use it!

The student did not swirl the flask appropriately and therefore stopped short of the endpoint. <em>FALSE. </em>The stopped short of the endpoint would explain a lower concentration of the real value.

The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water. <em>TRUE. </em>If you don't rinse the buret with KMnO₄ solution the concentration of this solution in the buret will be lower than real concentration doing you spend more titrant volume overestimating the amount of H₂O₂ moles.

The student added an extra 15 mL of distilled water to the H₂O₂ solution. <em>FALSE. </em>The addition of water doesn't change the amount of H₂O₂ moles in the solution.

I hope it helps!

6 0
3 years ago
77.33 grams of aluminum is heated from 155°C to 175°C. The specific heat of aluminum is 0.921
Dmitry [639]

Answer:

1424

Explanation:

Q=mco

Qis energy

M is mass

C is specific energy

O is temperature change

Substitute all the information you have in the equation then you will get the answer

5 0
3 years ago
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