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3 years ago

7

A welding torch produces a flame by burning acetylene fuel in the presence of oxygen the energy released from the acetylene fuel

is then used to melt a metal which energy transformation happens during this process

2 answers:

Volgvan3 years ago

8 0

Answer : Chemical energy to thermal energy

Explanation : When the welding torch uses the acetylene fuel for producing flame it is using the chemical acetylene to generate energy in form of light, after the flame is produced the fuel is used for melting a metal which is utilizing the thermal energy. So the ultimate final product is generated from chemical to thermal and so the energy transformation is from chemical energy to thermal energy.

enot [183]3 years ago

8 0

Answer:

A. Chemical energy turns into thermal energy.

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Would you weigh more or less on the moon than you do on Earth, and why?

VladimirAG [237]

Answer:You would weigh less on the moon because there is less gravity on the moon.

Explanation:

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3 years ago

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A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t

Margarita [4]

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

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3 years ago

Which is the correct formula for the compound made when aqueous solutions containing a dissolved magnesium compound and a dissol

jenyasd209 [6]

It would be MgCl₂ ~ Magnesium Chloride.

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4 years ago

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In the following equation:

Luda [366]

Answer:

FeCl₃

Explanation:

4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂

Given => 7moles 9moles

A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.

4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂

Given => 7/4 = 1.75* 9/3 = 3

*Smaller value => FeCl₃ is limiting reactant.

NOTE: However, when working problems, one must use original mole values given.

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3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago