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Salsk061 [2.6K]
3 years ago
11

How many grams of NH3 can be dissolved in 50 grams of water at 50oC?

Chemistry
1 answer:
weqwewe [10]3 years ago
5 0

15 grams of  NH3 can be dissolved

<h3>Further explanation</h3>

Given

50 grams of water at 50°C

Required

mass of NH3

Solution

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility  

  • 1. Temperature:
  • 2. Surface area:
  • 3. Solvent type:
  • 4. Stirring process:

We can use solubility chart (attached) to determine the solubility of NH3 at 50°C

From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g

So that the solubility in 50 grams of water is:

= 50/100 x 30

= 15 grams

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Answer:

appears to be moving back toward unmanned exploration in the form of deep-space satellite exploration.

Explanation:

NASA moved from unmanned probing of space in its initial stage of space exploration to manned space exploration and moon landing and now has gone back to unmanned exploration into deeper space. Deep space is not completely known yet, and the exploration take year and are probably very dangerous for manned exploration for now.

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Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion.
raketka [301]

Explanation:

The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.

This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose H^{+} ion.

Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.

Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.

       HOI < HOBr_{2} < HOCl_{3} < HOF

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3 years ago
List in ascending order the different layers of atmosphere, their heights, and main their features. Help please!!! :•)
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6 0
3 years ago
To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water
Masteriza [31]

Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

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Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

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4 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

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Putting values in above equation, we get:

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By Stoichiometry of the reaction:

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So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

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Putting values in above equation, we get:

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