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Nina [5.8K]
3 years ago
8

8. What is the molarity of a CsOH solution if 30.0 mL of the solution is neutralized by 26.4 mL 0.25 M HBr solution (Hint: Ma x

Va = Mb x Vb. Rearrange to solve for Mb)?
Ma = Molarity of Acid (M – mol/L)
Va = Volume of Acid (mL needs to be converted to L)
Mb = Molarity of Base (M – mol/L)
Vb = Volume of Base (mL needs to be converted to L)
CsOH = Base = What’s the molarity of the Base? = Mb
HBr = Acid = What’s the molarity of the acid? = Ma
Does the problem give you a volume for the acid? = Va
Does the problem give you a volume for the base? = Vb

HBr + CsOH = CsBr + H2O 
Chemistry
2 answers:
anyanavicka [17]3 years ago
6 0

Answer:

Remember, Molarity is moles of solute divided by Liters of solution. ... Hint: Don't forget to convert mL to L and use the formula Ma x Va = Mb x Vb. You are ... 8 What is the molarity of a CsOH solution if 30.0 mL of the solution is neutralized by 26.4 mL 0.25 M HBr solution ( Hint: Ma x Va = Mb x Vb. Rearrange to solve for Mb )? ..

Explanation:

pls follow me

sergey [27]3 years ago
6 0

Answer:

0.22M

Explanation:

The equation for the reaction is given below:

HBr + CsOH —> CsBr + H2O

From the above, the mole ratio of the acid to base is 1.

Data obtained from the question include:

Ma = 0.25M

Va = 26.4mL

Mb =?

Vb = 30mL

Ma x Va = Mb x Vb

Mb = (Ma x Va) / Vb

Mb = ( 0.25x26.4 ) / 30

Mb = 0.22M

Therefore, the molarity of CsOH is 0.22M

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The units of density are kg/m2. If the density of a liquid is 760.0 kg/m' what is the specific volume? a) 1.316 x 10 m2/kg b) 1.
zlopas [31]

Answer:

The correct answer is: 1.316 . 10⁻³ m³/kg.

Explanation:

The density (ρ) of a substance is the ratio of its <em>mass (m)</em> to its <em>volume (V)</em>. At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.

\rho = \frac{m}{V}

The specific volume (ν) of a substance is the ratio of its <em>mass</em> to its <em>volume</em>. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.

\nu = \frac{V}{m}=\frac{1}{\rho }

Given we know the density of the liquid, we can use this relationship to find out its specific volume:

\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg

6 0
3 years ago
Pleaseeeeeee hlepThe half-life of cobalt-60 is 5.26 years. After 10.52 years, 5 grams of a 20-gram sample will remain.
vladimir1956 [14]

The half-life of cobalt-60 is 5.26 years. After 10.52 years, 5 grams of a 20-gram sample will remain is TRUE

<u>Explanation:</u>

Mass of cobalt = 20 g  

Half-life = 5.26 years  

Mass remains after 10.52 years = 5 g  

This can be solved by using given below formula, m(t)=m_{o}\left(\frac{1}{2}\right)^{\frac{l}{5.26}}

m_{0} = initial mass  

t = number of years from when the mass was m_0  

m(t) = remaining mass after t years  

Number of half-lives = \frac{\text { Time elapsed }}{\text { Half -life }}

Number of half-lives = \frac{10.52 \text { years }}{5.26 \text { years }}

Number of half-lives = 2  

At time zero = 20 g  

At first half-life = \frac{20\ g}{2}  = 10 g  

At second half life = \frac{10\g}{2} = 5 g  

The given statement is true.

4 0
3 years ago
If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was
ANEK [815]

Answer:

Approximately 64\%.

Explanation:

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

The actual yield of \rm O_2 was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g).

Note the ratio between the coefficient of \rm H_2O\, (g) and \rm O_2\, (g):

\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}.

This ratio will be useful for finding the theoretical yield of \rm O_2\, (g).

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

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Calculate the formula mass of \rm H_2O and \rm O_2:

M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}.

M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}.

Calculate the number of moles of molecules in 9.8\; \rm g of \rm H_2O:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}.

Make use of the ratio \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2} to find the theoretical yield of \rm O_2 (in terms of number of moles of molecules.)

\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})}  \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}.

Calculate the mass of that approximately 0.271996\; \rm mol of \rm O_2 (theoretical yield.)

\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}.

That would correspond to the theoretical yield of \rm O_2 (in term of the mass of the product.)

Given that the actual yield is 5.6\; \rm g, calculate the percentage yield:

\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}.

4 0
3 years ago
Practice Explaining Heat Flow
zloy xaker [14]

Answer:

The right choice is "burning gas → pan → water → eggs"

Explanation:

Heat will flow from the body of higher temperature to the cooler body.

So, heat will firstly flow from burning gas to pan by conduction then to the water in contact with pan then to the egg present inside the water.

<u>So, the right choice is :</u>

"burning gas → pan → water → eggs"

6 0
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