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IrinaK [193]
4 years ago
6

Consider two different types of molecules reacting in the gas phase. which change below will not cause an increase in the rate o

f the reaction? select one:
a. add a catalyst.
b. increase the temperature at constant volume.
c. increase the volume at constant temperature.
d. add more molecules of each type to the system at constant volume.
Chemistry
1 answer:
kenny6666 [7]4 years ago
7 0
Correct answer is
<span>c. increase the volume at constant temperature.</span>
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Question 6
Hoochie [10]

Answer:

Explanation:

Area of a circular hubcap is 615 square inches

Area of a circular shaped object is given by :

A=\pi r^2

r is radius of hubcap.

If the radius of the hubcap were quadrupled, it means r' = 4r. New area will become:

\dfrac{A}{A'}=\dfrac{r^2}{r'^2}

A' is new area

A'=A\times \dfrac{r'^2}{r^2}\\\\A'=615\times \dfrac{(4r)^2}{r^2}\\\\A'=615\times 16\\\\A'=9840\ \text{inches}^2

New area of the hubcap is 9840 square inches.

6 0
3 years ago
What is the conjugate base of HSO−4 ?
Setler79 [48]
Although it has a negative charge, it will never accept a
H
+
to form
H
2
S
O
4
(sulfuric acid) . That is because sulfuric acid is a strong acid and completely disassociates in water.
Therefore, the sulfate ion (
S
O
2
−
4
) is the conjugate base of
H
S
O
−
4
.
8 0
1 year ago
while doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm^3. the accepted value for the density
vagabundo [1.1K]

Answer:

Percent error = 5.6%

Explanation:

Given data:

Measured value of density by students = 2.85 g/cm³

Accepted density of aluminium = 2.70 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [ Measured value - accepted value / accepted value ]× 100

Now we will put the values in formula:

Percent error = [ 2.85 g/cm³  - 2.70 g/cm³ / 2.70 g/cm³ ]× 100

Percent error = [0.15/2.70 g/cm³ ] × 100

Percent error = 0.056× 100

Percent error = 5.6%

7 0
4 years ago
10.9010 has how many significant figures
inna [77]

It has six significant figures. You count the 0s that come at the end because it shows you it wasn't rounded.

5 0
3 years ago
Read 2 more answers
Write a balanced molecular and net ionic equation for the following reactions
Natali [406]

Answer:

a. Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\

b. Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)

Explanation:

Hello,

a. In this case, the overall reaction is:

Al(OH)_3(s)+3HBr(aq)\rightarrow AlBr_3(aq)+3H_2O

Nevertheless, the ionic version is:

Al(OH)_3(s)+3H^++3Br^-(aq)\rightarrow Al^{3+}+3Br^-(aq)+3H_2O(l)\\

Since the base is insoluble, thereby, the balanced net ionic equation turns out:

Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\

Since bromide ions become spectator ions.

b) In this case, the overall reaction is:

Pb(NO_3)_2(aq)+2LiI(aq)\rightarrow PbI_2(s)+2LiNO_3(aq)

Nevertheless, the ionic version is:

Pb^{2+}(aq)+2(NO_3)^-(aq)+2Li^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2Li^+(aq)+2(NO_3)^-(aq)

Since lead (II) iodide is insoluble whereas lithium nitrate does not, thereby, the net ionic equation turns out:

Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)

Since lithium and nitrate ions become spectator ions.

Regards.

3 0
4 years ago
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