What volume does 8.5 g NH₃ occupy at STP? a. 2.81 L
b. 5.61 L
c. 11.21 L
d. 22.41 L
e. 44.81 L

Answers

Answer 1

Answer: First, convert grams to moles by dividing 8.5 by the molar mass of NH₃ (N=14,H=1)
14+1+1+1=17
8.5/17=.5 moles NH₃
Multiply .5 by the conversion factor of moles to L (22.4)
.5*22.4=11.2 L NH₃
11.2 is just 11.21 rounded down, so your answer is C) 11.21 L

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Which equation best represents the process of photosynthesis ? A. 6CO2+12H2O-> light+C6H12O6+6O2+6H2O B.6CO2+12H2O+light->C6H12O6+6O2+6H2O C.C6H12O6+6O2+6H2O+light->6CO2+12H2O D.C6H12O6+6O2+6H2O->light+6CO2+12H2O

Answers

The correct answer is B. The equation that represents the process of photosynthesis is: 6CO2+12H2O+light->C6H12O6+6O2+6H2O. Photosynthesis is the process in plants to make their food. This involves the use carbon dioxide to react with water and make sugar or glucose as the main product and oxygen as a by-product.

Answer: The correct answer is Option B.

Explanation:

Photosynthesis is defined as a process in which a plat uses carbon dioxide reacts with water in the presence of sunlight to produce glucose as a product, oxygen and water as a byproduct.

The chemical equation for this reaction follows:

$6CO_2+12H_2O+\text{light}\rightarrow C_6H_(12)O_6+6O_2+6H_2O$

By Stoichiometry of the reaction:

6 moles of carbon dioxide reacts with 12 moles of water in the presence of light to produce 1 mole of glucose, 6 moles of oxygen and 6 moles of water.

Hence, the correct answer is Option B.

What is an endothermic reaction?

Answers

Endothermic reactions are reactions that take in energy. This is usually in the form of heat. An example could be evaporation because heat is being absorbed by water and molecules are moving away from each other.

A reaction where energy is taken in from the surroundings causing temperature to decrease. Bond breaking requires energy and bond making releases energy. Therefore, if energy is taken in from the environment then in that reaction, more bonds are broken than formed. e.g. melting an ice cube.

The standard enthalpy change for the reaction of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH∘ = -227.8 kJ . Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO4(aq) (in kilojoules per mole). [For H2O(l),ΔH∘f = -285.8kJ/mol].

Answers

Answer:

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

Which of the following best explains why most chemical reactions proceed more quickly when the concentrations of reactants are increased?a. The increased concentration increases the number of collisions between molecules.
b. The products of chemical reactions are more stable at higher concentrations.
c. At higher concentrations, reactant molecules move more quickly.
d. At higher concentrations, product molecules are able to catalyze the reaction.

Answers

that is definitely not an A and C , so It'll be a B , most probably .

Which electrons are removed first when forming cations of Period 4 transition elements?

Answers

Electrons are always removed from the valence electron shells and from the valence orbitals within the shell when forming a cation. For example, Cr has a electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d5 4s1
When forming the ion Cr3+, the new electronic configuration will be 1s2 2s2 2p6 3s2 3p6 3d3. We can see that 3 electrons are removed from the valence shell following the order of the highest energy orbitals to lower energy.

Which of the following is a molecular compound?F) NO2
G) Ca(NO3)2
H) H2SO4
J) NaOH