How many moles are in 68.16 grams of NH3 ?

What happens when Ethane is treated with bayers reagent

swat32

Explanation:

यह लेख एक आधार है। जानकारी जोड़कर इसे बढ़ाने में विकिपीडिया की मदद करें।

उस पदार्थ या यौगिक को अभिकर्मक (reagent) कहते हैं जो किसी तंत्र में रासायनिक अभिक्रिया उत्पन्न करने के लिये डाला या मिलाया जाता है। उस पदार्थ को भी अभिकर्मक कहेंगे जिसे यह जांचने के लिये मिलाया जाता है कि कोई अभिक्रिया होती है या नहीं। इस तरह के कुछ वैश्लेषिक अभिकर्मक हैं - फेहलिंग का अभिकर्मक (Fehling's reagent), मिलॉन का अभिकर्मक (Millon's reagent) तथा टॉलीन का अभिकर्मक (Tollens' reagent)।

6 0

3 years ago

Which formula represents 1,2-ethanediol?

Oliga [24]

Answer:

1) C2H4(OH)2

Explanation:

A 1,2-ethanediol has an ethane structure consisting of two Carbon atoms with a hydrogen from each carbon substituted by a hydroxyl group. This makes it a 1,2-diol.

8 0

2 years ago

Omg I’ve been stressing for the past minutes can someone please help me

Alona [7]

Answer:

I think it would be the last answer

Explanation:

5 0

3 years ago

Read 2 more answers

How do you convert from grabs to moles of a substance

Lorico [155]

The correct question is as follows:

How do you convert from grams to moles of a substance

1. Divide by the molar mass

2. Subtract the molar mass

3. Add the molar mass

4. Multiple by the molar mass

Answer: In order to convert from grams to moles of a substance divide by the molar mass.

Explanation:

The number of moles of a substance is the mass of substance in grams divided by its molar mass.

The formula to calculate moles is as follows.

$Moles = \frac{mass}{molar mass}$

This means that grams are converted to moles when grams is divided by molar mass.

Thus, we can conclude that in order to convert from grams to moles of a substance divide by the molar mass.

8 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago