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3 years ago

Which of the following is not a characteristic of life?

Chemistry

1 answer:

Sophie [7]3 years ago

7 0

Answer:

Explanation:

Well movement is, reproduction is, responsiveness is, metabolism is, so accumulation would be the answer.

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The following is an example of which type of boundary?

kow [346]

I believe that it most likely would be D.

8 0

3 years ago

What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?

Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

So, the mass of 575 mL is 575 g = 0.575 kg.

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.36 m) = - 4.39 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.

∴ The freezing point of the solution is - 4.39 °C.

6 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

An element's atomic mass does not include the mass of its Electrons Protons Neutrons

klio [65]

Mass of electrons are not included when calculating the atomic mass of element. Atomic mass of the element is equal to proton + neurons. Example the the mass number of Nitrogen can be calculated as 7 protons + 7 neutrons to give 14 . thus nitrogen has a mass number of 14

4 0

3 years ago

Calcular la cantidad de NaOH necesaria para preparar medio litro de disolución 2,5 N. (Dato: peso molecular del NaOH = 40 g/mol)

kolbaska11 [484]

Answer:

The amount of NaOH required to prepare a solution of 2.5N NaOH.

The molecular mass of NaOH is 40.0g/mol.

Explanation:

Since,

NaOH has only one replaceable -OH group.

So, its acidity is one.

Hence,

The molecular mass of NaOH =its equivalent mass

Normality formula can be written as: $Normality=\frac{mass of solute NaOH}{its equivalent mass} * \frac{1}{volume of solution in L} \\$

Substitute the given values in this formula to get the mass of NaOH required.

$2.5N=\frac{mass of NaOH}{40g/mol} *\frac{1}{1L} \\mass of NaOH=2.5N*40gmol\\ = 100.0g$

Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g

8 0

2 years ago