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3 years ago

JJ thomson experiment the atomic theory?

Chemistry

1 answer:

olga nikolaevna [1]3 years ago

3 0

Cathode Ray experiment to figure out electrons exist -> atoms can be broken up into subatomic particles

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What is a way of transforming energy by exerting a force over distance?

xz_007 [3.2K]

Answer:

a. Work

Explanation:

If you apply a force over a given distance - you have done work. Work = Change in Energy. If an object's kinetic energy or gravitational potential energy changes, then work is done. The force can act in the same direction of motion.

8 0

3 years ago

Read 2 more answers

If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need

dimulka [17.4K]

In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps

5 0

3 years ago

A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3

ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

$V_1=250ml$

$V_2=?$

$P_{total}=735 mmhg$

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

$P_{water}=42.2mmhg$

$T_1=35°C=35+273=308 K$

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

$P_{gas}=P_{total}-P_{water}$

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

$\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}$

$V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1$

$V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1$

Here $P_2 \:and\: T_2$ are standard pressure and temperature respectively.

we have

$P_2=750mmhg \:and\: T_2=273K$

Substituting value, we get

$V_2=\frac{692.8}{750}*\frac{273}{308} *250$

$V_2= 819.51 ml$

4 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

3 years ago

Balancing Equations. How many of the element should I drag to the center?

babunello [35]

Answer:

4FeS + 7O₂ ----> 2Fe₂O₃ + 4SO₂

Explanation:

You have to drag the elements shown on the right under the formula. For example, you would have to drag 4 of the FeS molecules under the 4FeS text in the formula. Then place 7 O₂ molecules under the 7O₂ text, etc.

8 0

3 years ago