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Alenkinab [10]
2 years ago
8

A stock solution of ammonia in water is 28 wt% NH3. The Formal Weight (FW) of NH3 is 17.03 g/mol). The solution density at room

temperature is 0.90 g/mL. What is the molarity of the solution in units of mol/L
Chemistry
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

[NH₃] = 14.7 mol/L

Explanation:

28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.

Let's determine the amount of ammonia:

28 g . 1 mol / 17.03g = 1.64 moles of NH₃

You need to consider that, when you have density's data it is always referred to solution:

Mass of solution is 100 g, let's find out the volume

0.90 g/mL = 100 g /V

V = 100 g / 0.90mL/g → 111.1 mL

We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L

mol/L = 1.64 mol/0.1111L → 14.7 M

mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution

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How many moles of sucrose (C12H22O11) would be in 8.7 L of a 1.1 M solution of sucrose?
poizon [28]

Answer:

9.57 mol.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

<em />

<em>M = (no. of moles of solute)/(V of the solution (L)).</em>

<em></em>

∴ M = (no. of moles of sucrose)/(V of the solution (L)).

1.1 M = (no. of moles of sucrose)/(8.7 L).

<em>∴ no. of moles of sucrose = (1.1 M)(8.7 L) = 9.57 mol.</em>

5 0
3 years ago
The reduction of nitrogen monoxide is described by the following chemical equation: 2H2 (g) +2NO (g) 2H20 ()N2 (g Suppose a two-
Ipatiy [6.2K]

Answer:

Reasonable Second step- N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

Explanation:

The given single step chemical reaction is as follows.

2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)

Suppose a two-step mechanism is proposed for this reaction,

The reaction occured in two steps they are as follows.

Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)

Step-2:N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

8 0
3 years ago
After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is
Tanzania [10]

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

<h3>Ksp = 8.8x10⁻⁵</h3>
4 0
3 years ago
Please answer this question ASAP
Nezavi [6.7K]

Answer:

put a test tube over the opening, remove it and quickly put a lit splint near the mout or in the tube. if you hear a squeaky pop it is hydrogen.

Explanation:

hydrogen ignites in air.

4 0
2 years ago
Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
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