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2 years ago

What is the ratio of molecules in the chemical equation below N2+3N2-> 2NH3

Chemistry

1 answer:

irina [24]2 years ago

8 0

If you had to choose one of these it is 4:2.

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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i

In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

Moles CaCl₂ = Moles CaCO₃:

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0

2 years ago

PLEASE HELP I DONT WANT TO FAIL!! ):

Fudgin [204]

Hope this helps :) I didn’t know how to write subscripts so I wrote it down on some paper.

3 0

3 years ago

Choose an organism that you are familiar with, and explain how the three parts of the cell theory relate to the organism

jek_recluse [69]

Organism: Apple
All cells from the apple come from cell division occurring from the apple in order for it to grow.
An apple has multiple cells
The small living part in an apple is a cell thus showing that cells are the basic living unit.

6 0

2 years ago

Read 2 more answers

What is the density of an object that has a volume of 34.2 cm^3 and a mass of 19.6 g? (a) 0.573 g/cm^3 (b) 1.74 g/cm^3 (c) 670 g

KengaRu [80]

Answer: The density of the object will be $0.573g/cm^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of object = 19.6 grams

Volume of object= $34.2cm^3$

Putting in the values we get:

$Density=\frac{19.6g}{34.2cm^3}=0.573g/cm^3$

Thus density of the object will be $0.573g/cm^3$

8 0

3 years ago

How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]

lisabon 2012 [21]

Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 = 110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce = $\frac{3}{2}$ x moles of KClO3

So, O2 = (3 / 2) x 0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = 273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L

3 0

3 years ago