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3 years ago

Using the formula weight of tris (fw = 121.1), how much (to the nearest 0.01 g) is needed to prepare 50 ml of a 0.50 m solution?

Chemistry

1 answer:

Katyanochek1 [597]3 years ago

8 0

Answer:

3.03 g

Solution:

Data Given:

Molarity = 0.50 mol.L⁻¹

M.Mass = 121.1 g.mol⁻¹

Volume = 50 mL = 0.05 L

Step 1: Calculate moles as,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting values,

Moles = 0.50 mol.L⁻¹ × 0.05 L

Moles = 0.025 mol

Step 2: Calculate Mass of Tris as,

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting values,

Mass = 0.025 mol × 121.1 g.mol⁻¹

Mass = 3.03 g

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Significant digits are the number of digits that reflect the precision of a measurement or number.

Delvig [45]

Answer:

True

Explanation:

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Numbers that do not contribute to the precision of a reading should not be counted as significant.

There are rules of assigning significant numbers:

Leading or trailing zeros are insignificant and should only be counted as a place holder.
All non-zero digits are significant
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3 years ago

what pressure is generated when 5 mol of ethane is stored in a volume of 8 dm 3 at 25°C? Base calculations on each of the follow

Goshia [24]

Answer:

a) 40,75 atm

b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

PV = nRT

where:

P = pressure [atm]
V = volume [L]
n = number of mole of gas [n]
R= gas constant = 0,08205 [atm.L/mol.°K]
T=absolute temperature [°K]

Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the intermolecular forces and the space occupied by the gas

$\frac{Pv}{RT} = 1 + \frac{B}{v}$

where:

v is the molar volume [L/mol]
B is the second virial coefficient [L/mol]
P the pressure [atm]
R the gas constant = 0,08205 [atm.L/mol.°K]

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

$\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}$

and v = 3 L/5 moles = 0,6 L/mol

$P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm$

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3 years ago

As an object falls freely near the earths surface , the loss in gravitational potential energy of the object is equal to its wha

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Increase in kinetic energy as well as energy loss to the surroundings in the form of heat ( negligible)

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The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago