1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Yuki888 [10]
4 years ago
5

When sunlight passes through a prism, the different wavelengths separate into a(n)________ of colors.

Chemistry
2 answers:
Masteriza [31]4 years ago
7 0
When sunlight passes through a prism, the different wavelengths separate into a spectrum of colors.
marishachu [46]4 years ago
3 0
Spectrum of colors. Hope this helped!
You might be interested in
Did you hear that Oxygen is going out with Magnesium?
Morgarella [4.7K]
Yes... And they had a child named magnesium oxide.
6 0
3 years ago
Read 2 more answers
How many atoms are in 1.4 mol of phosphorus trifluoride (PF3)?
joja [24]

Number of atoms in 1.4 mol of Phosphorus trifluoride (PF₃) : 8.428 x 10²³

<h3>Further explanation  </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

1.4 mol of Phosphorus trifluoride (PF₃), number of atoms :

\tt N=1.4\times 6.02\times 10^{23}\\\\N=8.428\times 10^{23}

5 0
3 years ago
The density of pure gold is 19.3g/cm3. What is the volume of 1.00 g of pure gold?
soldier1979 [14.2K]

Answer: Volume of 1g of pure gold v=0.05181 \mathrm{cm}^{3}

Given;

Density of a pure gold=19.3 g / c m^{3}

Mass of a pure gold =1g

To find:

Volume of 1g of pure gold

Solution:

According to the formula,

Density= Mass/Volume

     \rho=m / v

Where \rho=density of pure gold

           m=mass of pure gold

          v=volume of pure gold

From the above equation volume can be calculated as

v=m / \rho

Substitute the values of mass and density value in the above equation

v=1 / 19.3

v=0.05181 \mathrm{cm}^{3}

Result:

Thus the volume of 1g of pure gold is v=0.05181 \mathrm{cm}^{3}

8 0
3 years ago
W + BgCz2 --&gt; WCz + Bg<br> Balance this equation
yanalaym [24]

Answer: W + BgCz2 arrow WCz + Bg

2 W + BgCz2 arrow  2 WCz + Bg

Explanation:

Cz has 2 so you balcne the other side of WCz.

Since you Balcanes the Cz you changed the W and you Balcanes the other W on the left side.

3 0
3 years ago
What is the oxidation number of chlorine in Al(ClO4)3? Please give me unique work! THANKSS
sveta [45]
It would be +7 oxidation number.
[All compounds are zero]
Al is in group III [ or 13 depending on the system ] so the oxidation state is 3.
You have 12 O at -2 each for-24. That leaves you at -24+3=-21. So, 3 Cl atoms must be +21 to balance the -21 there which makes +21/3 or +7 for each Cl.
3 0
3 years ago
Other questions:
  • How much heat, in kJ, is required to raise the temperature of 50 g of water by 5.53°C? (Round to the nearest 10 kJ, and enter on
    7·1 answer
  • What are the four symbols for physical states of reactants and products?
    8·1 answer
  • The tendency of a moving object to continue moving in a straight line or a stationary object to remain in place is called
    9·2 answers
  • A student conducting the iodine clock experiment accidentally makes an S2O32- stock solution that is too concentrated. How will
    8·1 answer
  • Which of the following is NOT a property of an acid?
    13·2 answers
  • Mutations are changes in the genes of living things. Mutations are common in all life forms and usually harmless. In fact, somet
    9·2 answers
  • What is the pH of a solution with a hydrogen concentration of 4.3 X 10^-12
    5·1 answer
  • The nervous system involves actions that control
    13·2 answers
  • Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction e
    6·1 answer
  • COURSES GROUPS RESOURCES GRADE REPORT Q OOO try: 913322 4 Honors Chemistry Tests/Quizzes ormulas Reactions Quiz Questic Question
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!