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Harman [31]
3 years ago
13

Can someone help please! Is it physical or chemical? Just say yes or no

Chemistry
2 answers:
Kamila [148]3 years ago
5 0

Answer:

yess I think it is bit then again I haven't really do t that

Svetach [21]3 years ago
4 0

Answer:

Physical Reaction

4.

8.

9.

11.

15.

17.

19.

20

Chemical Reaction

2.  Hydrochloric acid reacts with  potassium....

3. Water is heated and changes..

5. potassium chlorate...

6.

7.

10.

12.

13.

14.

16.

18.

The numbers are the question number

the number are under their specified places if it is chemical then it should be under chemical change and if it is physical it should be under physical change

Question Number 1. is not written as i don't know the answer.

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Hiii please help me for balancing chemical equations:
Zarrin [17]

Answer:

KI + Cl2 = KCl + I2

Explanation:

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I think number 9 is D and number 10 is A

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A student pipets 10 ml of a 0.0693 m solution of kbr into a 500 ml volumetric flask and dilutes to the mark. what is the concent
Sonja [21]

The initial concentration of solution is 0.0693 M. The volume of solution taken is 10 mL and it is diluted to a final volume of 500 mL.

According to dilution law, the product of initial concentration and volume is equal to the product of final concentration and volume as follows:

C_{1}V_{1}=C_{2}V_{2}

Here, C_{1} is initial concentration, C_{2} is final concentration, V_{1} is initial volume and V_{2} is final volume.

Rearranging to calculate final concentration,

C_{2}=\frac{C_{1}\times V_{1}}{V_{2}}

Putting the values,

C_{2}=\frac{0.0693 M\times 10 mL}{500 mL}=0.001386 M

Therefore, concentration of the resulting solution is 0.001386 M.

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3 years ago
Tin hydrogenooxolate formula
vladimir2022 [97]

Answer:

Tin(IV) Hydrogen Oxalate. Alias: Stannic Hydrogen Oxalate. Formula: Sn(HC2O4)4. Molar Mass: 474.8178. :: Chemistry Applications:: Chemical Elements, Periodic Table.

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alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima
svetlana [45]

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  = 0.10 M

The chemical equation for this reaction is :

\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = \mathtt{ \dfrac{0.035  \ mol}{ 1.0  \  L }} = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  \mathtt{CH_3COOH \ \ \   +  \ \ \ \ OH^-  \ \ \to \ \ CH_3COO^-  \ \ \ + \ \ \  H_2O}

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log \mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}

The pH of this solution = 4.74 + log \mathtt{\dfrac{0.135}{0.065}}

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

7 0
3 years ago
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