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3 years ago

A conducting sphere has a net charge of 4.8  1017

1 answer:

5 0

A net charge of the sphere is the overall charge of the sphere. If it says that there are excess electrons, it means that the net charge is negative. Also, the quantity of charge of sub particles are very small so that must be 10 to the power of negative 17. So, I think the net charge is -4.8×10⁻¹⁷ Coulombs.

Then, we use the electrical charge of an electron which is equal to 1.60217662×10⁻¹⁹ C/electron. Therefore,

-4.8×10⁻¹⁷ Coulombs * 1 electron/-1.60217662×10⁻¹⁹ C = 299.592

Therefore, there are approximately 300 excess electrons.

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HELP.....From the information presented in the activity series of metals below (4 metals included), answer the following questio

san4es73 [151]

Answer:

a)there would be no reaction

Explanation:

The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.

In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.

Would aluminum replace magnesium to form a new compound or would there be no reaction?

Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.

6 0

3 years ago

An unknown metal sample has a mass : of 11.29. and a volume of 2 cm3. What is the : density of the unknown metal?

tatuchka [14]

Answer:

$\boxed{\sf Density \ of \ the \ unknown \ metal = 5.645 \ g/cm^3}$

Given:

Mass = 11.29 g

Volume = 2 cm³

To Find:

Density of unknown metal

Explanation:

Formula:

$\boxed{ \bold{Density = \frac{Mass}{Volume}}}$

Substituting value of mass & volume in the formula:

$\sf \implies Density = \frac{11.29}{2} \\ \\ \sf \implies Density = 5.645 \: g/cm^3$

3 0

3 years ago

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

3 years ago

Read 2 more answers

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

5 0

3 years ago

HELP PLZ ITS DUE SOON!!

sergey [27]

It is the second option! the square with more orange dots will have to move over so that there is an equal amount or equilibrium! another way to look at it is, if we just moved the one dot on the left side then that side would be empty and there would not be equilibrium

7 0

3 years ago