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2 years ago

How is electricity and magnetism related

1 answer:

3 0

Electricity and Magnetism are related as electricity can actually make magnetic material (iron, nickel or silver) magnetic. This is called an electro magnet (a temporary magnet created by wrapping coils around a magnetic core (the electricity powers the coils, magnetising the material)).

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Which planet do scientists believe prevented the asteroids in the asteroid belt from forming a planet? A. Earth B. Jupiter C. Ma

True [87]

I woud say B because jupiter has more of a gravitational pull

5 0

2 years ago

Read 2 more answers

How are Kelvin scale and celcius scale similar <br>

Alekssandra [29.7K]

Answer: The Kelvin scale is related to the Celsius scale. The difference between the freezing and boiling points of water is 100 degrees in each, so that the kelvin has the same magnitude as the degree Celsius.

Explanation:

Celsius is, or relates to, the Celsius temperature scale (previously known as the centigrade scale). The degree Celsius (symbol: °C) can refer to a specific temperature on the Celsius scale as well as serve as a unit increment to indicate a temperature interval(a difference between two temperatures or an uncertainty). “Celsius” is named after the Swedish astronomer Anders Celsius (1701-1744), who developed a similar temperature scale two years before his death.

K = °C + 273.15

°C = K − 273.15

Until 1954, 0 °C on the Celsius scale was defined as the melting point of ice and 100 °C was defined as the boiling point of water under a pressure of one standard atmosphere; this close equivalence is taught in schools today. However, the unit “degree Celsius” and the Celsius scale are currently, by international agreement, defined by two different points: absolute zero, and the triple point of specially prepared water. This definition also precisely relates the Celsius scale to the Kelvin scale, which is the SI base unit of temperature (symbol: K). Absolute zero—the temperature at which nothing could be colder and no heat energy remains in a substance—is defined as being precisely 0 K and −273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C.

7 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:

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