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faust18 [17]
3 years ago
7

How many grams of oxygen would be needed to completely react with 18 grams of methane (CH4)?

Chemistry
1 answer:
S_A_V [24]3 years ago
6 0

Answer:

35.91 grams of oxygen

Explanation:

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what is the effect of using higher concentration of sodium hydrogen carbonate on the rate of reaction?
Alex
Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.

Reaction rate increases with concentration, as described by the rate law and explained by collision theory. As reactant concentration increases, the frequency of collision increases. The rate of gaseous reactions increases with pressure, which is, in fact, equivalent to an increase in concentration of the gas.
7 0
2 years ago
Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360
Aleks [24]

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.  

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

3 0
3 years ago
Read 2 more answers
How could we increase the strength of a solenoid?
12345 [234]

Strength of the magnetic field can be increased by:

1) increasing the current in the coil.

2) increasing the number of coils in the solenoid; and.

3) using a soft iron core within the solenoid.

Hope this helps!

8 0
3 years ago
At a given temperature, 4.06 atm of H2 and 3.5 atm of Cl2 are mixed and allowed to come to equilibrium. The equilibrium pressure
Llana [10]

<u>Answer:</u> The value of K_p for the given chemical reaction is 0.1415

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as K_p

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_p is written as:

K_p=\frac{p_{C}^cp_{D}^d}{p_{A}^ap_{B}^b}

For the given chemical equation:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)

The expression for K_p for the following equation is:

K_p=\frac{(p_{HCl})^2}{(p_{H_2)}(p_{Cl_2})}

We are given:

p_{HCl}=1.418atm\\p_{H_2}=4.06atm\\p_{Cl_2}=3.5atm

Putting values in above equation, we get:

K_p=\frac{(1.418)^2}{(4.06)\times (3.5)}\\\\K_p=0.1415

The value of K_p for the given chemical reaction is 0.1415

5 0
3 years ago
Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,
erma4kov [3.2K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

          <u>Lue</u>  <u>Arg</u>  <u>Lys</u>  <u>Arg</u>  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the  polypeptides contains 8 amino acids

   From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

     Also from the question we are told that the Carboxypeptidase  released Ser this mean the the C-terminal  amino acid would be Ser ( i.e Serine)

   The Amino acids would in the polypeptides would be arranged like this  

         

         <u>Lue</u> __  __  __  __  __ __  <u>Ser</u>

Now from the question we are told that treatment with cranogen bromide form two peptides.

   Now generally  cranogen bromide divides a peptide on  the C- side(i.e the extreme left side ) of  Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

   So this means that any peptides  containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)  

    Thus the fifth amino acid is  Met  

So the sequence of these amino acid would now be  

          <u>Lue</u> __  __  __  <u>Met</u>  __ __  <u>Ser</u>

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides  

   Now generally Trypsin divides a peptide  on the  C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys  must have them as their C- terminal  amino acids

   From the first peptide in the two peptide formed we see that  Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

       <u>Lue</u> __  __  __  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

Looking  at the first amino acid formed from the Trypsin-catalyzed hydrolysis  we see that Arg would be the fourth amino acid of  octapeptide      as Trypsin divides a peptide  on the  C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the  Trypsin-catalyzed hydrolysis  we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the  C- side of  Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

          <u>Lue</u>  <u>Arg</u>  <u>Lys</u>  <u>Arg</u>  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

           

4 0
3 years ago
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