In a laboratory investigation, a student separates

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

6 0

3 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

Answer: The energy released in the given nuclear reaction is 1.3106 MeV.

Explanation:

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

3 years ago

. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 0

3 years ago

Identify the formula for an alkene.

ioda

Answer:

CnH2n-2

im pretty sure thats the answer

5 0

3 years ago

An atom in an excited state contains more of what type of energy than the same atom in the ground state

LenaWriter [7]

Answer:

Total Energy of the electrons

Explanation:

An atom in an excited state is when the total energy of electrons can be lowered by transferring molecule(s) over to different orbitals. in an excited state atom, not all electrons are at the lowest energy levels

4 0

3 years ago