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2 years ago

The handle of tin-snips (metal cutter) are made long.give reason

Chemistry

2 answers:

NeX [460]2 years ago

4 0

Answer:

Because it makes the cutting process more easier.

Explanation:

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masha68 [24]2 years ago

3 0

Answer:

Because in metal cutter we need large we need large mechanical advantage to overcome the resistive load while cutting while cutting the metal sheets ,so that it can be handle easily by us.

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A cylinder contains 4.0 g of nitrogen gas. A piston compresses the gas to half its initial volume. Afterward, by what factor has

notka56 [123]

Answer:

The mass density will be doubled

Explanation:

Density is given by dividing the mass of a substance by its volume.
An increase in mass causes an increase in density and vice versa, while a decrease in volume causes an increase in density and volume.
Therefore, when the volume is halved, then the density will be doubled if the mass is kept constant.
This has no effect on the number of moles as the mass is constant.

8 0

2 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

2 years ago

2KOH+H2SO4=k2SO4+2H2O Is a balanced equation, displaying the combination of potassium hydroxide with sulfuric acid increase pota

Lerok [7]

Answer:

The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH

Explanation:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.

Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;

8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O

8 moles of KOH is required to make 4 moles of K₂SO₄.

6 0

2 years ago

his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo

Alinara [238K]

Answer:

$\%\ Composition\ of\ iron=69.92\ \%$

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is $Cr(NO_3)_3$

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

$\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100$

$\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100$

$\%\ Composition\ of\ iron=69.92\ \%$

5 0

2 years ago

The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG

Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate

ΔG∘ = -12.6 KJ/mole = -12600 J/mole

ΔG = -0.1 KJ/mole = -100 J/mole

[Creatine phosphate] = 25 mM = 25*10^-3 M

[Creatine] = 17 mM = 17*10^-3 M

[ATP] =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600 = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500 = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc = 6.44*10^-3

But,

Qc = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3 = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

= 2.1896 *10^-5 M

= 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0

2 years ago

The handle of tin-snips (metal cutter) are made long.give reason ​

The handle of tin-snips (metal cutter) are made long.give reason