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3 years ago

Once the torch is lit why must the acetylene flow be increased until the flame stops smoking

1 answer:

5 0

Once the torch is lit, the acetylene flow must be increased until the flame stops smoking before the oxygen is turned on for adjustment in order to keep the tip of the torch cool.

You should also note that while lighting the torch, you should keep the spark lighter near the tip but not covering it.

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50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4

VikaD [51]

Answer:

2.1 M is the molarity of the HCl solution.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

Molarity of HCl solution = $M_1=?$

Volume of HCl solution = $V_1=50.0mL$

Ionizable hydrogen ions in HCl = $n_1=1$

Molarity of NaOH solution = $M_2=1.05 M$

Volume of NaOH solution = $V_2=100.0 mL$

Ionizable hydroxide ions in NaOH = $n_2=1$

$n_1M_1V_1=n_2M_2V_2$ (neutralization )

$M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}$

$M_1=2.1 M$

2.1 M is the molarity of the HCl solution.

3 0

3 years ago

Examples of physical and chemical weathering

lorasvet [3.4K]

Mass is always conserved in a physical change. Energy may be released or absorbed when a substance changes from one physical state to another. In a chemical change, a chemical reaction yields a completely new substance. A substance's particles are changed during a chemical reaction.

5 0

3 years ago

Which list shows the phases of matter in order from the greatest average kinetic energy to the least average kinetic energy per

avanturin [10]

Answer:

Explanation:

3 0

2 years ago

How do I "predict the colour change that would be observed if carbon dioxide gas were bubbled into a mixture of calcium oxide, b

Effectus [21]

Hey there!

Just say what color you think the mixture would look like if those elements were combined. Personally I don't know because I don't have context, but if it comes to it just pick a color :)

Good luck, have a good night.

7 0

3 years ago

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago