All categories

2 years ago

Write ashort note on thefirst battle ofof panipat

Chemistry

1 answer:

mel-nik [20]2 years ago

5 0

Answer:

The First Battle of Panipat was fought between the invading forces of Babur and the Lodi Empire, which took place on 21 April 1526 in North India. It marked the beginning of the Mughal Empire. This was one of the earliest battles involving gunpowderfirearms and field artillery.

You might be interested in

5 points

Orlov [11]

Answer:

The correct option is (a).

Explanation:

We need to find the product of given numbers.

First number = 29.5

Second number = 240

Product of two numbers = 29.5 × 240

= $\dfrac{295}{10}\times 240$

Zeroes from numerator and denominator get cancelled.

So,

$=295\times 24\\\\=7080$

Hence, the correct option is (a).

7 0

3 years ago

What do The seven dots around fluorine represent

borishaifa [10]

The seven dots around flourine are the valence electrons

7 0

3 years ago

How many iron atoms are in the formula for iron(III) oxide?

velikii [3]

Answer:

There is two iron atoms.

Explanation:

The formula iron oxide is Fe2O3 F e 2 O 3 therefore there would be two. Hope this helps. :)

8 0

3 years ago

Fabric is made of ___

GuDViN [60]

Answer:

Fabric is made of cotton.

Explanation:

Hope that helps :)

6 0

3 years ago

Read 2 more answers

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

3 0

3 years ago

Write ashort note on thefirst battle ofof panipat​

Write ashort note on thefirst battle ofof panipat