1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Jet001 [13]
3 years ago
15

What are some of the problems with using flame tests as a way to identify substances?

Chemistry
1 answer:
ASHA 777 [7]3 years ago
3 0
Different substances can produce similar results in color. This can cause <span>uncertainty with identifying the substance.</span>
You might be interested in
Consider the following reaction:
Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

                        SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)

Initial conc.    2.4\times 10^{-2}          0             0

At eqm.          (2.4\times 10^{-2}-x)   x              x

As we are given:

Concentration of Cl_2 at equilibrium = 1.3\times 10^{-2}M

That means,

x=1.3\times 10^{-2}M

The expression for equilibrium constant is:

K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}

Now put all the given values in this expression, we get:

K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}

K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}

K_c=0.0154

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0
3 years ago
1. How many joules of heat are required to raise the temperature of 750 g of water from 11.0 oC to 19.0 oC?
Leya [2.2K]

Answer:

  1. 25080 J
  2. 146.9 g
  3. 92.58 °C
  4. 0.808 J/g°C
  5. 117.09 g
  6. a. 1708.8 kJ  b.1246.56 kJ
  7. 368.55 kJ
  8. 6.81 kJ
  9. 5.50 grams of methane produces more heat than 5.5 grams of propane.

Explanation:

  1. The specific heat capacity of water=4.18 J/gK

The enthalpy change is calculated using the formula: ΔH=MC∅ where ΔH is the change in enthalpy, M the mass of the substance, C the specific heat capacity of the substance and ∅ the temperature change.

Thus, ΔH= 750g × 4.18 J/gK × (19-11)K

=25080 J

2. Enthalpy change= mass of substance × specific heat capacity of the substance× Change in temperature.

ΔH= MC∅

M= ΔH/(C∅)

Substituting for the values in the question.

M=8750 J/(0.9025/g°C×66.0 °C)

=146.9 grams

3. Enthalpy change =mass × specific heat capacity × Temperature

ΔH= MC∅

∅ = ΔH/(MC)

=6500 J/(250 g × 4.18 J/g°C)

=6.22° C

Final temperature =98.8 °C - 6.22°C

=92.58 °C

4. Specific heat capacity =mass × specific heat capacity × Temperature change.

ΔH=MC∅

C= ΔH/(M∅)

Substituting with the values in the question.

C = 4786 J/(89.0 g×(89.5° C-23°C))

=0.808 J/g°C

5. Heat lost lost copper is equal to the heat gained by water.

ΔH(copper)= ΔH(water)

MC∅(copper)=MC∅(water)

M×0.385 J/g°C× (75.6°C- (19.1 °C+5.5°C))=100.0g×4.18 J/g°C×5.5 °C

M=(100.0g×4.18J/g°C×5.5°C)/(0.385 J/g°C×51 °C)

=117.09 grams.

6 (a). From the equation 1 mole of methane gives out 890.4 kJ

There fore 2 moles give:

(2×890.4)/1= 1780.8 kJ  

(b) 22.4 g of methane.

Number of moles= mass/ RFM

RFM=12 + 4×1

=16

No. of moles =22.4 g/16g/mol

=1.4 moles

Therefore 1.4 moles produce:

1.4 moles × 890.4 kJ/mol=

=1246.56 kJ

7. From the equation, 2 moles of aluminium react with ammonium nitrate to produce 2030 kJ

Number of moles = mass/RAM

Therefore 9.75 grams = (9.75/26.982) moles of aluminium.

=0.3613 moles.

If 2 moles produce 2030 kJ, then 0.3613 moles produce:

(0.3631 moles×2030 kJ)/2

=368.55 kJ

8. From the equation, 4 moles of ammonia react with excess oxygen to produce 905.4 kJ of energy.

Number of moles= mass/molar mass

RMM= 14+3×1= 17

Therefore 0.5113 grams of ammonia = (0.5113 g/17g/mole) moles

= 0.0301 moles

If 4 moles produce 905.4 kJ, then 0.0301 moles produce:

(0.0301 moles×905.4 kJ)/4 moles

=6.81 kJ

9. From the equations, one mole of methane produces 890 kJ of energy while one mole of propane produces 2043 kJ.

Lets change 5.5 grams into moles of either alkane.

Number of moles= Mass/RMM

For propane, number of moles= 5.5g/ 44.097g/mol

=0.125 moles

For methane number of moles =5.5 g/ 16g/mol

=0.344 moles

0.125 moles of propane produce:

0.125 moles×2043 kJ/mol

=255.375kJ

0.344 moles of methane produce:

0.344 moles× 890 kJ/mol

= 306.16kJ

Therefore, 5.5 grams of methane produces more heat than 5.5 grams of propane.

6 0
3 years ago
When solid ammonium chloride dissociates at a certain temperature in a 0.500 dm3 container, ammonia and hydrogen chloride are fo
SIZIF [17.4K]
1) Reaction

<span>NH4Cl(s) ---> NH3(g) + HCl(g)


2) equilibrium equation, Kc


Kc = [NH3] * [HCl]


3) Table of equilibrium formation



step               concentrations
                       </span>
<span>                            NH4Cl(s)                     NH3(g)             HCl(g)


start                      1.000 mole                     0                    0


react                        - x


produce                                                       +x                  + x
                         ------------------                ----------              -----------

end                         1 - x                             +x                    +x
            

1 - x = 0.3 => x = 1 - 0.3 = 0.7


[NH3] = [HCl] = 0.7/0.5 liter = 1.4                (I used 0.500 dm^3 = 0.5 liter)


4) Equilibrium equation:


Kc = [NH3] [HCl] = (1.4)^2  = 1.96


Which is the number that you were looking for.


Answer: Kc = 1.96
</span>
8 0
3 years ago
Consider a reactant with an order of 1: What effect does that reactant have on the rate equation when the concentration is doubl
____ [38]

Answer:

1) The reaction rate is double with respect to that reactant

Explanation:

Hello,

By considering the rate law:

-r_A=kC_A

If we double the reactant A concentration, by definition, the rate will be doubled as well since the C_A power is one (order 1), this could be proved just by checking it out in the rate law.

Best regards.

6 0
3 years ago
Reading the above graph to the nearest tenth of a milliliter, what is the volume of sodium hydroxide at the equivalence point?
Vlada [557]

Answer:

The volume of sodium hydroxide at the equivalence point is:

  • <u>14.9 mL of sodium hydroxide</u>.

Explanation:

<u>The equivalence point occurs when, in this case, the HCl is completely neutralized with the solution of NaOH, how you can see this doesn't occur in the last point but occurs in the nineteenth point, where the pH is no more acid (below to 7) but is 11 approximately</u>, then you must see in the X-axis from this point and you can see the volume is almost 15, by this reason I calculate the valor of 14.9 milliliters.

7 0
3 years ago
Other questions:
  • BRAINLIESTTT ASAP!!! PLEASE HELP ME :)
    15·1 answer
  • The synthesis of dipropyl ether can be accomplished using 1-propanol. What reactants and conditions are necessary for this to oc
    15·1 answer
  • 15. A food sample contains 25 g carbohydrate, 8 g fat and 5 g protein. The energy content is:
    8·1 answer
  • 1. Determine the molarity given 0.6762 mol KOH and a volume of<br> 500.00 mL.
    12·1 answer
  • How do I find the chemical formula?
    8·1 answer
  • In the electron cloud model, the electron cloud is denser in some locations than in others. What do the denser areas represent?
    6·2 answers
  • Metallic behavior is generally associated with elements with partially filled p orbitals. elements with unpaired electrons. elem
    5·1 answer
  • Why do convection currents occur?
    6·1 answer
  • At 45.0 C and a pressure of 9.9 kPa a sample of gas has a volume of 1.033 L. If the pressure is increased to 1245 kPa what will
    14·1 answer
  • How many moles of hydrogen atoms are<br> present in 15 moles of C2H6O?<br> Answer in units of mol.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!