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3 years ago

6

Which tool is used to hold workpieces tightly so that both of your hands can be free to work on them? A. Snap-ring pliers B. Nee

dlenose pliers C. Vise D. Bench

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

A vice, you can tighten it and then work on what is held in it without having to hold it still or in the position you want.

Natasha_Volkova [10]3 years ago

6 0

C. A view to hold it tight so you can work on it!

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This is the SI unit for force meter liter Newton m/s/s

morpeh [17]

The newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared. Newton's second law of motion states: F = ma, multiplying m (kg) by a (m/s 2 ).

I don't understand your question, but I think that would help.

4 0

3 years ago

How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?

Sophie [7]

Answer:

The correct answer is 0.4 L.

Explanation:

The mentioned question can be solved by using the equation,

C = K × Pgas--------(i)

Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.

Now to find the volume of water, the formula to be used is,

PV = nRT-----------(ii)

Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.

PgasV = nRT

Pgas = nRT/Vgas

The value of Pgas is inserted in equation (i) we get,

C = K × nRT/Vgas

It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.

n/Vliquid = K × nRT/Vgas

1/Vliquid = KRT/Vgas

Vliquid = Vgas/KRT--------------(iii)

Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.

Now putting the values in equation (iii) we get,

Vliquid = 1.52 L / 0.158 × 0.082 × 294

Vliquid = 1.52 / 3.809

Vliquid = 0.399 or 0.4 L

Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.

8 0

3 years ago

The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________

PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>

<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg</span>

<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.</span>

6 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

The percent composition by mass of nitrogen in NH OH (gram-formula mass = 35 grams/mole) is equal to

Temka [501]

Answer:

the percentage composition of mass of nitrogen in NH OH is 42.86 %

7 0

3 years ago