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ASHA 777 [7]
3 years ago
13

Amines, amides, and amino acids are organic compounds containing the element:

Chemistry
2 answers:
3241004551 [841]3 years ago
8 0
They all contain the same element
Colt1911 [192]3 years ago
5 0
They contain Carbon, Nitrogen, Hydrogen, Oxygen, and Sulfur
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What is the maximum amount of kcl that can dissolve in 200g of water?
Likurg_2 [28]
The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.
8 0
2 years ago
Question 2: you have two systems: 1) solid gold and water 2) solid sodium and waterWhich one can spontaneously release energy an
Valentin [98]

Hey there!

Solid Sodium and water will react spontaneously and release energy.  This is based on the reactivity series. Sodium is a highly reactive metal and hence, it is placed at the top of the reactivity series. This is because it loses its outermost electron very readily. When it comes in contact with water, it reacts with it violently to form sodium hydroxide and hydrogen gas. This reaction is exothermic and hence, accompanied with a release of energy.  Gold lies at the bottom of the reactivity series as it is very stable and does not give away its outermost electrons easily. Therefore, when it comes in contact with water, there is no reaction and no release of energy.

7 0
3 years ago
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

           [A]₀ is the initial concentration of A

           k is the rate constant, and

           t is the time

We also know that for a first order reaction

           k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

thus:

(pPH₃ )t  = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0
3 years ago
Could someone help me with this?
Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

  • Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

  • x is the oxidation number

#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

7 0
2 years ago
It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min
kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 8.1\times 10^{-2} min⁻¹

Initial concentration [A_0] = 0.1 M

Final concentration [A_t] = 1.0\times 10^{-2} M

Time = ?

Applying in the above equation, we get that:-

1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}

0.1e^{-8.1\times \:10^{-2}t}=10^{-2}

e^{-8.1\times \:10^{-2}t}=\frac{1}{10}

\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)

t=28.43\ min

3 0
3 years ago
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