What is the modern atomic model sometimes called?

The change of state from solid to gas without an intermediate liquid state is called evaporation. true and false

-BARSIC- [3]

It is false, evaporation is when water vapor heats and goes from a liquid to a gas. The change of state from solid directly to gas is called sublimation

8 0

3 years ago

The green light emitted by a stoplight has a wavelength of 505 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).

Kitty [74]

Answer:

5.9405*10^14 Hz

Explanation:

Remember the equation c=v*λ

v=frequency

λ=wavelength

By manipulating the equation, we can get it in terms of frequency:

c/λ=v

Now convert wavelength to meters (1nm=m*10^-9)

5.05*10^-7m (2 decimal places were shifted to make the number between 1-10, hence the -7 exponent)

Use the manipulated equation:

(3.00*10^8)/(5.05*10^-7), which yields:

5.9405*10^14 Hz (or s^-1)

6 0

3 years ago

Based on the activity series provided, which reactants will form products? F > Cl > Br > I CuI2 Br2 Right arrow. Cl2 Al

AleksAgata [21]

The reactants based on the reactivity capable of forming the products are $\rm CuI_2\;and\;Br_2$ . Thus, option A is correct.

The reactivity series has been the arrangement of the elements based on their reactivity. The highly reactive element has been capable of replacing the less reactive element in the compound, and form the product.

In the following reactions, the new product has been formed by:

$\rm CuI_2\;and\;Br_2$

From the reactivity series, Br has been more reactive than I, and thereby replaces iodine in the compound for the formation of product.

$\rm Cl_2\;and\;AlF_3$

The compound has been formed with F. Cl has been less reactive than F, and will not be able to form the product.

$\rm Br_2\;and\;NaCl$

The compound has been formed with Cl. Br has been less reactive than Cl, thus will not displaces Cl for the formation of new product.

$\rm CuF_2\;and\;I_2$

The compound has been formed with F. I has been less reactive than F, and will not be able to form the product.

The reactants that will be able to form the products have been $\rm CuI_2\;and\;Br_2$ . Thus, option A is correct.

For more information about the reactivity series, refer to the link:

brainly.com/question/10443908

5 0

3 years ago

892.970 cm2 to four significant figures

Sophie [7]

Answer:

Explanation:

There is a trick to this. Counting out the 4 places is easy enough.

Start at the 8 and count to your right. You land on the nine, but that is not quite your answer.

The 7 influences the 9 and the nine gets bumped up which makes it a 10. but you just record the zero part. The 1 gets added to the 2.

That in turn influences the 2 which makes it a three. So your answer is

893.0

The trick is that you must include the 0. You are guaranteeing that you know the accuracy of the fourth place. Weird but true.

It helps just slight to convert this to scientific notation.

8.9297 * 10^2 = 8.930 * 10^2

You get the same rounding problem as discussed above, but it might be a little easier to see.

You get the same answer and you have to go through the same steps.

5 0

3 years ago

How to balance this by oxidation state change method? . KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O

Juliette [100K]

In this redox reaction, we must identify which is reduced and which is oxidized. Cl is oxidized from -1 to 0, Mn is reduced from +7 to +2.

MnO−4+Cl−→Mn2++Cl2Balance the number of Cl first. MnO−4+2Cl−→Mn2++Cl2
MnO−4+10Cl−→2Mn2++5Cl2

Balance the number of oxygen by adding H2O 2MnO−4+10Cl−→2Mn2++5Cl2+8H2O

Balance the number of H by adding H+ 2MnO−4+10Cl−+16H+→2Mn2++5Cl2+8H2O

 Now, we know that there are 2 H+ in H2SO4, so, we can get 2MnO−4+10Cl−+8H2SO4→2Mn2++5Cl2+8H2O

 Put SO2−4 with Mn2+ 2MnO−4+10Cl−+8H2SO4→2MnSO4+5Cl2+8H2O

 Add K in front of MnO−4 and Cl− and balance it on the right 2KMnO4+10KCl+8H2SO4→12K++2MnSO4+5Cl2 + 8H20

The final answer is 2KMnO4+10KCl+8H2SO4→6K2SO4+2MnSO4+5Cl2+8H2O

6 0

4 years ago