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2 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Chemistry

1 answer:

Helen [10]2 years ago

5 0

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

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Given that the frequency of a wave is 9.12x10^-12 Hz, what must be the

polet [3.4K]

Answer:

The wavelength of wave is 0.33 ×10²⁰ m.

Explanation:

Given data:

Frequency of wave = 9.12×10⁻¹² Hz

Wavelength of wave = ?

Solution:

Formula:

Speed of light = wavelength × frequency

c = λ × f

λ = c/f

This formula shows that both are inversely related to each other.

The speed of light is 3×10⁸ m/s

Frequency is taken in Hz.

It is the number of oscillations, wave of light make in one second.

Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.

Now we will put the values in formula.

λ = 3×10⁸ m/s / 9.12×10⁻¹² Hz

Hz = s⁻¹

λ = 0.33 ×10²⁰ m

The wavelength of radiation is 0.33 ×10²⁰ m .

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3 years ago

It takes one oxygen to combine with two hydrogen atoms to form water. What is the mass ratio of oxygen to hydrogen? 1 to 2 4 to

Rasek [7]

Answer:

8 to 1.

Explanation:

Oxygen combines with hydrogen atoms to form water according to the balanced equation:

O₂ + 2H₂ → 2H₂O.

It is clear that one mole of oxygen combines with two moles of hydrogen atoms to form 2 moles of water.

So, the molar ratio of oxygen to hydrogen is (1 to 2).

The mass of 1 mole of oxygen = (no. of moles)(molar mass) = (1 mol)(32.0 g/mol) = 32.0 g.

The mass of 2 moles of hydrogen = (no. of moles)(molar mass) = (2 mol)(2.0 g/mol) = 4.0 g.

So, the mass ratio of oxygen to hydrogen (32.0 g/4.0 g) = (8: 1).

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3 years ago

Of the following, which is NOT a way to classify a biome?

slavikrds [6]

Answer:

Explanation:

Economic principles has nothing to do with biome determination.

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2 years ago

Read 2 more answers

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

3 years ago

When ammonium chloride crystals are dissolved in water, the temperature of the water decreases. What does the temperature change

sertanlavr [38]

Answer:

See Explanation

Explanation:

A decreasing temperature indicates that the dissolution process for the ammonium chloride requires input of energy from surroundings. That is, the process is essentially 2 parts => system (object of interest - NH₄Cl) and the surroundings (everything else - solvent - H₂O). The surroundings (water) solvent is showing a measured decrease in temperature or loss of energy (exothermic to surroundings) which flows into the system (NH₄Cl) and effects dissolution of salt into solution (endothermic to system).

3 0

3 years ago