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Delvig [45]
3 years ago
11

A rotameter calibration curve (flow rate versus float position) obtained using a liquid is mistakenly used to measure a gas flow

rate. Would you expect the gas rate determined in this manner to be too high or too low?
Chemistry
1 answer:
Veronika [31]3 years ago
5 0

Answer:

I would expect the gas rate determined in this manner to be too low

Explanation:

A Rotameter can be designed to respond to the sensitivity of density, velocity, to measure the flow rate of liquid or gas enclosed in a tube. Liquids are denser than gas, and since the gas rate to be determined needed to respond to the velocity head alone of the rotameter so as to bring the forces in the tube equilibrium. Knowing if there is no flow, then the float would remain at the bottom, so gas has to flow at a higher rate compared to the liquid so the float would be in a similar position making it easier to measure the flowrate. This leaves the gas rate to be determined too low.

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32. In general, ionization energy increases from left to right across a given
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What is law of conservation of mass​
dexar [7]

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Mass in an isolated system can neither be destroyed nor created.

Explanation:

6 0
3 years ago
Balance the following redox equations by the half-reaction method:
frozen [14]

Answer:

a)  Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2-  → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+   → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+  → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

<em>(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)</em>

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O  →  2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2  → MnO2 + 2H2O  + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

<em>(b) Bi(OH)3 + SnO2^2-  → SnO3^2- + Bi (in basic solution)</em>

Step 1: The half reactions

Reduction:  Bi(OH)3 + 3e-  → Bi

Oxidation : Sno2^2-  → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e-  → Bi + 3OH-)

3* (Sno2^2- +2OH-  → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2-  → 2Bi + 3SnO3^2- + 3H2O

<em>(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e-  → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e-  +14H+  → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

<em>(d) ClO3^- + Cl^− </em>→<em> Cl^2 + ClO^2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

                      ClO3^- + e- → ClO2

 2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

                   Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+   → Cl2 + 2H2O +2ClO2

<em>(e) Mn^2 + BiO3^− </em>→<em> Bi^3 + MnO^4− (in acidic solution)</em>

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+  → 5Bi^3+ + 7H2O + 2MnO4^-

6 0
4 years ago
mickey reacts magnesium with 1.0 M of hydrochloric acid. the reaction tales 2 minutes and 34 seconds to complete. what are 2 way
lutik1710 [3]

Answer:

Explanation:

To slow down this reaction, we can use any of the methods listed below:

  • Increase the size of the magnesium by using solid lumps of  the metal. This will take a much faster time to react than powered and granulated magnesium.
  • Reduce the concentration of the acid.
  • Let the reaction take place at a much lower temperature than that given.

These conditions will slow down a chemical reaction.

7 0
3 years ago
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