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2 years ago

Which of the following is the energy of moving electrons?

Chemistry

1 answer:

mezya [45]2 years ago

6 0

Answer:

Electrical energy

Explanation:

Electrical energy is caused by the moving of electrons.

hope it helps!

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Which of the following describes a way in which chemistry affects how you

stepladder [879]

Answer:

I guess A, I am not sure...

6 0

2 years ago

Read 2 more answers

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Pie

Answer:

$m = \frac{E}{c^2}$

$c = \sqrt{\frac{E}{m} }$

$E = 12.8 J$

Explanation:

Part 1: Solving for m

We are given that:

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

$m = \frac{E}{c^2}$

Part 2: Solving for c

We are given that:

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

$c^2 = \frac{E}{m}\\ \\ c=\sqrt{\frac{E}{m}}$

Part 3: Solving for E

We are given that:

m = 80 and c = 0.4

To get the value of E, we will simply substitute in the given equation:

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

4 0

2 years ago

A sphere with radius 1 m has temperature 18°C. It lies inside a concentric sphere with radius 2 m and temperature 25°C. The temp

gladu [14]

Answer:

The expression will be T=7r

Explanation:

From equation S=dT/dr, we can solve this equation by integrating it and usig the values 25C and 18C for T and 2 and 1 for the radius. The solution to the integration will result in the value for S. For this example, it will be 7. Therefore, the main expression will be T= 7r in which the temperature is in fuction of the radius between the spheres.

4 0

2 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

2 years ago

muriatic acid is the commercial name for hydrochloric acid that can be purchased from hardware stores as a solution that is 30%(

SSSSS [86.1K]

We have 30% (w/w) HCl this means :
100 g solution contains 30 g HCl
?? g solution contains 7.5 g HCl
By cross multiplication:
Mass of solution = $\frac{100 x 7.5 }{30}$ = 25 g solution

5 0

2 years ago