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kolbaska11 [484]
3 years ago
8

Determine whether or not each mixture is a buffer. check all that apply. check all that apply. koh and nh3 hbr and nacl hcl and

hbr hcho2 and nacho2
Chemistry
1 answer:
erastova [34]3 years ago
3 0
A buffer is usually composed of either:
- a weak base combined with its conjugate acid, or
- a weak acid combined with its conjugate base

Now, examining each of the choices:
1- KOH and NH3: since they are both bases, thus this mixture is not a buffer
2- HBr and NaCl: since HBr is a strong acid while NaCl is the salt of another different acid, thus this mixture is not a buffer
3- HCl and HBr: these are both considered to be strong acid, thus this mixture is not a buffer
4- HCHO2 and NaCHO2: as for HCH02, it is considered a weak base while NaCHO2 is considered its conjugate base, thus this mixture is a buffer

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Draw a mechanism for the reaction of methylamine with propanoic acid. In the box to the left, draw any necessary curved arrows.
miv72 [106K]

Answer:

Explanation:

find the solution below

8 0
3 years ago
Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn=
IgorC [24]

Answer: \Delta H^0=+0.3kJ/mol.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

S_{rhombic}+O_2(g)\rightarrow SO_2(g)    \Delta H^0_1=-296.06kJ   (1)

S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ  (2)

The final reaction is:  

S_{rhombic}\rightarrow S_{monoclinic}  \Delta H^0_3=?   (3)

By subtracting (1) and (2)

\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ

3 0
3 years ago
How many grams of a 5.5% solution can be prepared from 25 kg KCl
AnnZ [28]

Answer:

The percentage concentration of any solution is most commonly expressed as mass percent:

Mass % of any component of the solution =

(Mass of the component in the solution / Total mass of the solution) x 100

Other methods are:

Volume percentage:

Volume % of a component =

(Volume of the component/Total volume of the solution) x 100

Mass by volume percentage:

It is the mass of solute dissolved in 100 mL of the solution.

i.e. Mass by Volume percentage =

(Mass of solute in grams/Volume of solution in mL) x 100

Here's a point to be kept in mind :

Whenever we say mass or volume of the solution, you need to add the respective masses and volumes of ALL the components of the solution. Do NOT commit the error of taking the mass or volume of only the solute or solvent in the denominators of the above expressions.

The concentration of a solution is most of the time expressed as the number of moles of solute present in 1 Liter of the solution (also called molarity )

(There are also other ways to express concentration. Please follow this link. )

EXAMPLE:

(a) If 25 moles of NaCl are present in 100 L of a solution wherein H2O is the solvent, then the concentration of the solution is

25

100

=

0.25

mol⋅L

−

1

.

(b) What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in enough water to make a total of 225 mL of solution?

Solution

Calculate the number of moles of solute present.

Moles of NaOH = 15.0 g NaOH ×

1

mol NaOH

40.00

g NaOH

= 0.375 mol NaOH

Calculate the number of litres of solution present.

Volume = 225 mL ×

1

L

1000

mL

= 0.225 L soln

Divide the number of moles of solute by the number of litres of solution.

Molarity =

0.375

mol

0.225

L

= 1.67 mol/L

Explanation:

7 0
4 years ago
3,3-dibromo-4-methylhex-1-yne​
mel-nik [20]

Explanation:

<h2>see the attachment. hope it will help you... </h2>

5 0
4 years ago
Complete and balance the molecular equation for the reaction of aqueous copper(II) chloride, CuCl2, and aqueous potassium phosph
aleksandrvk [35]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which do not get involved in the chemical equation. It is also defined as the ions which are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of copper (II) chloride and potassium phosphate is given as:

3CuCl_2(aq.)+2K_3PO_4(aq.)\rightarrow Cu_3(PO_4)_2(s)+6KCl(aq.)

Ionic form of the above equation follows:

3Cu^{2+}(aq.)+6Cl^-(aq.)+6K^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Cu_3(PO_4)_2(s)+6K^+(aq.)+6Cl^-(aq.)

As, potassium and chloride ions are present on both the sides of the reaction, thus, they will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

3Cu^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Cu_3(PO_4)_2(s)

Hence, the net ionic equation is given above.

7 0
3 years ago
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