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3 years ago

The diagram below shows the positions of the Earth, Moon, and Sun. What is the date?

Chemistry

2 answers:

Anna35 [415]3 years ago

5 0

Answer is: B) June 21.
This is summer solstice or midsummer.
Summer solstice occurs when a planet's geographical pole on Northern Hemisphere is most inclined toward the Sun. Earth's maximum axial tilt toward the Sun is 23.44° and the Sun reaches its highest altitude of the year on the summer solstice.

fomenos3 years ago

3 0

Answer:

Explanation:

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Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

Dmitrij [34]

Answer :

(a) The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

The spectator ions are, $NH_4^{+}\text{ and }SO_4^{2-}$

(b) The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

(c) The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)$

The ionic equation in separated aqueous solution will be,

$2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)$

In this equation, $NH_4^{+}\text{ and }SO_4^{2-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

(b) The given balanced ionic equation will be,

$Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

(c) The given balanced ionic equation will be,

$Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)$

The ionic equation in separated aqueous solution will be,

$Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

3 0

3 years ago

Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?

pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0

3 years ago

A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$