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Karolina [17]
3 years ago
14

List three facts or accomplishments of Modern India (India today)​

Chemistry
1 answer:
viva [34]3 years ago
8 0

Answer:

India is one of the oldest countries in the world and is found in the Asian continent of the world.

They are known as the most populous country in the world.

They also have a greater linguistic diversity than any other large country. Lots of languages are available and accepted in the country.

They are very vast in Agriculture and accounts for 40percent of Mango production in the world

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2. The coefficients represent to molar ratios in a balanced equation.
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Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.
Jlenok [28]

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

6 0
3 years ago
The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal
rosijanka [135]

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -511.3kJ/mol=-511300.0J/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of \Delta n_gRT = 0

R = gas constant = 8.314 J/mol.K

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above formula, we get

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K

\Delta U=-511300.0J/mol+4955.144J/mol

\Delta U=-506344.856J/mol=-506.3kJ/mol

Therefore, the internal energy change is -506.3 kJ/mol

6 0
3 years ago
4.50 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1
Anna11 [10]

Answer:

\mathbf{C_{10}H_8}   ( Naphthalene )

Explanation:

Given that:

4.50 g of a Compound X is made up of Carbon , Hydron and Oxygen

It's molecular molar mass = 128 g/mol

Compound X undergoes combustion reaction and the product yield :

CO_2 with mass 15.47g and :

H_2O with mass 2.53 g

The objective is to use this information to determine the molecular formula of X.

We all know that ; number of moles = mass/molar mass

where the molar mass of H_2O is 18 g/mol

number of moles of H_2O product = 2.53 g/18 g/mol

number of moles of H_2O product = 0.1406 moles

Also; the molar mass of CO_2 product = 44 g/mol

number of moles of CO_2 product = 15.47g/ 44 g/mol

number of moles of CO_2 product =  0.3516 moles

number of moles of Compound X in the reactant side= 4.50 g /128 g/mol

number of moles of Compound X n the reactant side= 0.03516 moles

Now; number number of moles of CO_2 in reactant = 0.3516 moles/0.03516 moles

Now; number number of moles of CO_2 in reactant = 10

number of moles of H_2O reactant = 0.1406 moles × 2/0.03516

number of moles of H_2O reactant = 7.997 ≅ 8

Since we said the Compound X is known to be made of Carbon C , Hydrogen H and Oxygen O

Then the molecular formula can be written as :

\mathbf{C_{10}H_8O_{x}}

In order to find the x; we have

128  = (12 × 10 + 1 × 8 + 16 × x)

128 = 120 + 8 + 16x)

128  =  128 + 16 x

128 - 128 = 16 x

0 = 16 x

x = 0/16

x = 0

As x = 0 ; hence there are no oxygen present in the reaction

Thus; the molecular formula of Compound X = \mathbf{C_{10}H_8} which is also known as Naphthalene

5 0
3 years ago
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