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3 years ago

What are the criteria for purity

Chemistry

1 answer:

Sauron [17]3 years ago

5 0

Answer:

melting and boiling points

Explanation:

The technique for determination of melting and boiling points will also be described in this unit. Pure solid and liquid compounds possess sharp melting and boiling points. Therefore, melting and boiling points of a compound

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Based on the definition of specific heat and the information provide below, if the same amount of heat is added to 1 kg of all t

REY [17]

It would be the one with the highest Specific Heat constant

The equation for specific heat is:

Q=mCT

If we rearrange for T, we get:

T=Q/mC

Since m is just 1 Kg, we can ignore it and say that:

T=Q/C

Let’s just give an arbitrary number to C and solve for each individual species

Q=1 J

Helium - 1J/5193 = .00019257

Steel - 1J/490 = .00204082

Aluminum - 1J/897 =.00111483

Water - 1J/4182 = .00023912

This proves that Steel, with the lowest Specific Heat will increase the most.

7 0

3 years ago

Read 2 more answers

What would be an advantage and a disadvantage to using litmus as a titration indicator???

lesantik [10]

To indicate whether the elements is basic or acid by changing of colour.(advantage)
can give out bad gas which can cause suffocate to the people.(disadvantage).

4 0

3 years ago

The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

Nikolay [14]

Answer:

$\Delta H_{rxn3}=-162.5 kJ/mol$

Explanation:

The reaction we need to calculate:

$O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)$

1) $O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)$

$\Delta H_{rxn}=-122.8 kJ/mol$

We need the ClO in the products side, so we use the inverse of this reaction:

$Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g)$

$\Delta H_{rxn1}=122.8 kJ/mol$

2) $2 O_3 (g) \longrightarrow 3 O_2 (g)$

$\Delta H_{rxn2}=-285.3 kJ/mol$

Now we need to combine this two:

$Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g)$

$Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g)$

The enthalpy of reaction:

$\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol$

$\Delta H_{rxn3}=-162.5 kJ/mol$

4 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago

Mr. Greenjeans wants to know which fertilizer helps grow the biggest pumpkins. He plants three pumpkin seeds of the same variety

andrew11 [14]

A)Mr.Greenjeans needs a control group for comparison.
hope this helps

3 0

4 years ago

Read 2 more answers