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3 years ago

According to the following position vs. Time graph the bicyclist was

Chemistry

1 answer:

julia-pushkina [17]3 years ago

3 0

Answer:

According to the following position vs. Time graph the bicyclist was : D

D. Not moving for the first two seconds, then begin moving at the constant speed.

Explanation:

$Speed = \frac{Distance}{time}$

Along Y-axis = Distance in meters

Along X-axis = Time taken in second

For first 2 seconds the The object has not changed its position.There is Zero distance covered.So , Speed = 0
After 2 second,The Object changed the distance by equal amount in equal interval of time.(Uniform Speed).Hence at each point after 2 second, there is same value speed

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How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

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3 years ago

describe the placement of the crucible lid on the crucible when heating the magnesium. Why is it important that this be done cor

Andreas93 [3]

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3 years ago

The standard enthalpy change for the following reaction is 873 kJ at 298 K.

Flura [38]

Answer: - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

$2KCl(s)\rightarrow 2K(s)+Cl_2(g)$ $\Delta H_1=873kJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s)$ $\Delta H_2=?$

According to the Hess’s law, if we divide the reaction by half then the $\Delta H$ will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value $\Delta H_2$ for the reaction will be:

$\Delta H_2=\frac{1}{2}\times (-873kJ)$

$\Delta H_2=-436.5kJ$

Hence, the value of $\Delta H_2$ for the reaction is -436.5 kJ.

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3 years ago

The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen

tino4ka555 [31]

Answer: The mass percentage of $Tl_2SO_4$ is 5.86%

Explanation:

To calculate the mass percentage of $Tl_2SO_4$ in the sample it is necessary to know the mass of the solute ( $Tl_2SO_4$ in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the $TlI$ precipitate. We can establish a relation between the mass of $TlI$ and $Tl_2SO_4$ using the stoichiometry of the compounds:

$moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.$

Since for every mole of Tl in $TlI$ there are two moles of Tl in $Tl_2SO_4$ , we have:

$moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol$

Using the molar mass of $Tl_2SO_4$ we have:

$mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g$

Finally, we can use the mass percentage formula:

$mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%$

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3 years ago

Hi, im 15 i live in indiana and i need friends please i love tacos and im an all around fun person but very emotionally drained.

Andrei [34K]

Answer:

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Explanation:

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3 years ago

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