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
svp [43]
3 years ago
12

Can rotation and revolution occur at the same time? Explain.

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

yes

Explanation:

Revolution is the movement of an object around another object. So, Earth revolves around the Sun, and the Moon revolves around Earth. At the same time, Earth and the Moon are also rotating. Earth's rate of revolution is about 365 days (one year), and its rate of rotation is about 24 hours (one day).

You might be interested in
If carbon has the electron configuration of (2,4) what would the electron configuration of the element to the right of it on the
Svetllana [295]

Answer:

2,5

Explanation:

When move from left to right in periodic table atomic number increases. So element to the right of carbon has one electron more than carbon therefore it's electronic configuration would be 2,5

3 0
3 years ago
Question 2 of 15
guapka [62]

Answer:

B. The products are nuclei of elements that are different from the original elements.

6 0
1 year ago
What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

8 0
3 years ago
Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi
ivann1987 [24]

Answer:The formulas of ionic compounds are:

a)CuBr

b)Mn_2O_3

c)Hg_2I_2

d)Mg_3(PO_4)_2

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains Cu^+ ion.

Cu^++Br^-\rightarrow CuBr

(b) Manganese oxide : Given that it contains Mn^{3+} ion.

Mn^{3+}+O^{2-}\rightarrow Mn_2O_3

(c)Mercury iodide :Given that it contains Hg_2^{2+}

Hg_2^{2+}+I^-\rightarrow Hg_2I_2

(d) Magnesium phosphate :Given that it contains PO_4^{3-}

Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2

4 0
3 years ago
A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

5 0
3 years ago
Other questions:
  • When a 7.00 g7.00 g sample of KBrKBr is dissolved in water in a calorimeter that has a total heat capacity of 2.72 kJ⋅K−1,2.72 k
    6·1 answer
  • Acetic acid has a boiling point of 118.5 C and a boiling point constant of 3.08. What is the boiling point of a 3.20 m solution
    10·1 answer
  • Which specific characteristic must a signaling molecule have in order to bind to a cytosolic or nuclear receptor?
    11·1 answer
  • HALPPPPPP MEH YALL<br> THANKSSSSSS
    6·2 answers
  • A small sack of sand has a density of 1.5 g/cm3 and a mass of 1500 g. How much space (volume) does the sand occupy? 750 cm3 2250
    10·2 answers
  • Using the balanced chemical equation below. 2Al2O3 --&gt; 4Al + 3O2 How many moles of oxygen are produced if 11.0 mol of Al are
    6·1 answer
  • Boiling Point temperatures are dependent
    6·1 answer
  • Pls help me. Show proof on why it’s that answer.
    9·1 answer
  • A gas takes up a volume of 17.0L, has a pressure of 2.3 atm, and a temperature of 299K. If I raise the temperature to 355K and l
    7·1 answer
  • QUESTION 14
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!