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3 years ago

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What is the most likely evidence of an energy transformation taking place when a skater glides over ice? A.Skating blades become

cold as kinetic energy is transformed into thermal energy B.Skating blades become cold as kinetic energy is transformed into gravitational energy C.Small tracks of water are left on the ice as kinetic energy is transformed into thermal energy D.Small tracks of water are left on the ice as kinetic energy is transformed into gravitational energy

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

The correct answer is C. The weight downward and the friction moving forward creates thermal energy through kinetics and melts a tiny bit of the ice making it slick enough for the skate to slide.

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In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

3 years ago

How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*

MrMuchimi

<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$ .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

$3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L$

Hence, 6.57 L of solution can be made.

5 0

3 years ago

0.00099989<br> 3 sig figs, scientific notation

mixas84 [53]

1.00*10^3
You’d need to lower the exponent because rounding to 3 sig figs changes the 9’s to - 1000. Keep the 0’s.

5 0

2 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

A total of 20.0 mL of sodium hydroxide (NaOH) was neutralized by 30.0 mL of 0.250 M hydrogen bromide (HBr). What was the concent

Karolina [17]

Answer:

0.375 M

Explanation:

NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)

Concetration of acid CA= 0.250M

Concentration of base CB= ????

Volume of acid VA= 30.0mL

Volume of base VB= 20.0mL

Number of moles of acid nA= 1

Number of moles of base nB= 1

CA VA/CB VB= nA/nB

CB= CAVAnB/VB nA

= 0.25× 30×1/20×1= 0.375 M

8 0

3 years ago