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viktelen [127]
3 years ago
10

What is the formula to use for this question, "how far does light travel in 2.5s"?

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0
The definition of the speed of light is exactly 299,792,458 meters per second, so to find how far it travels in a time period, multiply the speed of light times the time. Aka c=299,792,458m/s where c is speed of light, m is meters, and s is seconds. So for example to find how far light travels in 5 seconds, multiply by 5.
You might be interested in
What is the name of this compound?
Digiron [165]
2,2-dichloroheptane  because we have two Cl we should use the prefix ''di'' and the numbers "2" and "2" indicated that the two functional groups are located on the same second carbon 
6 0
3 years ago
How many atoms are in 3.20 moles of carbon
DIA [1.3K]

Answer:

19.264×10^{23} atoms are present in 3.2 moles of carbon.

Explanation:

It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.

So, as 1 mole of any element = Avagadro's number of atoms = 6.02×10^{23} atoms

It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.

As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.

1 mole of carbon = 6.02 ×10^{23} atoms

Then, 3.20 moles of carbon = 3.20 × 6.02 ×10^{23} atoms

Thus, 19.264×10^{23} atoms are present in 3.2 moles of carbon.

6 0
2 years ago
What is the boiling point elevation constant, Kb, of diethyl ether if 38.2 g of the nonelectrolyte benzophenone, C6H5COC6H5, dis
kakasveta [241]

Answer: the boiling point elevation constant is 1.73^0C/m

Explanation:

Elevation in boiling point is given by:

\Delta T_b=i\times K_b\times m

\Delta T_b=T_b-T_b^0= = Elevation in boling point

i= vant hoff factor = 1 (for non electrolyte)

K_b =boiling point constant = ?

m= molality

\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (diethylether)= 330 g = 0.33 kg

Molar mass of solute (benzophenone)= 182 g/mol

Mass of solute (benzophenone) = 38.2 g

(35.7-34.6)^0C=1\times K_b\times \frac{38.2g}{182g/mol\times 0.33kg}

K_b=1.73^0C/m

Thus the boiling point elevation constant is 1.73^0C/m

3 0
2 years ago
50 points I need help on this whole work sheet about converting moles
kozerog [31]

Answer:

Explanation:

11)

Answer:

9.08 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of O₂ = 22.7 mol

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                               O₂      :        P₂O₅

                                 5      :          2

                                 22.7  :        2/5×22.7 = 9.08

12)

Answer:

7 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of P = 14 mol

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with P.

                               P        :        P₂O₅

                               4        :          2

                                14      :        2/4×14 = 7

13)

Answer:

76.25 mol

Given data:

Number of moles of P =  61 mol

Number of moles of O₂ react = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P with O₂.

                                  P         :        O₂

                                  4          :          5

                                 61          :        5/4×61 = 76.25

14)

Answer:

1.25 mol

Given data:

Number of moles of P₂O₅ = 0.5 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                                P₂O₅         :        O₂

                                  2            :          5

                                0.5          :        5/2×0.5 = 1.25

15)

Answer:

20 mol

Given data:

Number of moles of P₂O₅ = 8 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                                P₂O₅       :        O₂

                                  2            :          5

                                  8            :       5/2×8 = 20

16)

Answer:

12 mol

Given data:

Number of moles of silver made = ?

Number of moles of Ag₂O = 6 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with Ag₂O .

                       Ag₂O      :       Ag

                           2         :        4

                           6          :        4/2×6 = 12

17)

Answer:

25 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 6.25 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with O₂ .

                          O₂             :       Ag

                           1               :        4

                           6.25          :      4×6.25 = 25

18)

Answer:

9.8 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 2.45 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with O₂ .

                          O₂             :       Ag

                           1               :        4

                          2.45          :      4×2.45 = 9.8

19)

Answer:

4.4 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 2.2 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

                           O₂            :       Ag₂O

                           1               :         2

                          2.2            :        2×2.2  = 4.4

20)

Answer:

1.5 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 0.75 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

                           O₂            :         Ag₂O

                           1               :            2

                          0.75            :        2×0.75 = 1.5

8 0
3 years ago
Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.
mars1129 [50]

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

7 0
2 years ago
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