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2 years ago

which event would be less widely visible from Earth: a partial lunar eclipse or a total lunar eclipse? explain please

Chemistry

1 answer:

adell [148]2 years ago

6 0

Answer:

I had the same question and I put a total lunar eclipse.

Explanation:

A total lunar eclipse would be less widely visible because, in a partial lunar eclipse the moon only has to partly be in the Earth's shadow.

I don't know if this is helpful or not, but this is what I put if you still needed it.

Still stuck? Get 1-on-1 help from an expert tutor now.

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Which of the following statements is true for an exothermic reaction

Georgia [21]

In an exothermic reaction the energy of the product is less than the energy of the reactants.

3 0

4 years ago

Read 2 more answers

A sample of Element X is found to contain 67.25% of isotope type 1 (85.91) and 32.75% of isotope type 2 (87.91). Calculate the a

KengaRu [80]

Answer:

The average atomic mass is 86.565

Explanation:

An element's average atomic mass is the sum of the products of the masses of the isotopes of the element and their percentage abundance divided by 100

The average atomic mass of the Element X is given as follows;

Isotope type 1 (85.91) ×

Isotope ${}$ Abundance (%) Mass /100 Portion of average mass

Type 1 ${}$ 67.25 × 85.91 ÷ 100 57.774475

Type 2 ${}$ 32.75 × 87.91 ÷ 100 28.790525

${}$ Average atomic mass = 86.565

The average atomic mass = 86.565.

5 0

3 years ago

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

<u>Answer:</u> The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?

Phantasy [73]

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x ( x molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241 ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

8 0

3 years ago

In the explosion of a hydrogen-filled balloon, 0.80 g of hydrogen reacted with 6.4 g of oxygen. How many grams of water vapor ar

mylen [45]

0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O

Explanation:

write the values given in the question

In an explosion of hydrogen balloon, 0.80g of hydrogen is reacted with 6.4g of oxygen.

Write the balanced equation

2 H2 + O2 ---> 2 H2O

0.80 g H2 x (1 mole / 2g) = 0.4 moles H2

6.4 g O2 x (1 mole / 32g) = 0.2 moles O2

from balanced equation, 2 moles H2 react with 1 mole O2

therefore

0.4 moles H2 x (1 mole O2 / 2 moles H2) = 0.2 moles O2

since we need 0.2 moles O2 to react with all the H2 and we have exactly that amount, the amounts are said to be

"stoichiometric" and either reactant can be considered limiting

from the balanced equation, 2 moles H2 --> 2 moles H2O.. therefore

0.4 moles H2 x (2 moles H2O / 2 moles H2) = 0.4 moles H2O

Then

0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O

8 0

3 years ago