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3 years ago

What evidence have you discovered to explain the impact of the Moon's tilted orbit on eclipse formation?

Chemistry

1 answer:

professor190 [17]3 years ago

7 0

It’s far away from us and there for that how the moons spin

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The three main geochemical cycles of Earth are _____, _____, and _____. These geochemical cycles continuously remove and add mat

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3 years ago

Read 2 more answers

Volume in liters of carbon dioxide produced at stp when 0.85 grams of butane reacts

MAVERICK [17]

Volume in liters of Carbon dioxide : 0.672

<h3>Further explanation</h3>

Reaction(combustion of butane-C₄H₁₀)

2C₄H₁₀+13O₂⇒8CO₂+10H₂O

mol butane (MW=58,12 g/mol) :

$\tt \dfrac{0.85}{58,12}=0.015$

mol CO₂ : mol C₄H₁₀ = 8 : 4, so mol CO₂ :

$\tt \dfrac{8}{4}\times 0.015=0.03$

At STP, 1 mol = 22.4 L, so volume CO₂ :

$\tt 0.03\times 22.4~L=0.672~L$

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3 years ago

In the valence bond theory, a bond forms between two atoms because

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4 years ago

Which Period 4 element has the most metallic properties?<br> (1) As (3) Ge<br> (2) Br (4) Sc

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The answer is (4) Sc. The metallic properties has the rule that from left to right, the metallic property is from high to low. Then from the position of these four elements, we can get the answer.

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3 years ago

Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from –10 °C to 182°C. Assume that the specif

TiliK225 [7]

Answer : The amount of heat required is, $1.16\times 10^6J$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(182^oC)$

The expression used will be:

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat required for the reaction

m = mass of ice = 346 g

$c_{p,s}$ = specific heat of solid water or ice = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.184J/g^oC$

$c_{p,g}$ = specific heat of gaseous water = $1.99J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $333J/g$

$\Delta H_{vap}$ = enthalpy change for vaporization = $2260J/g$

Now put all the given values in the above expression, we get:

$\Delta H=[346g\times 2.09J/g^oC\times (0-(-80))^oC]+346g\times 333J/g+[346g\times 4.184J/g^oC\times (100-0)^oC]+346g\times 2260J/g+[346g\times 1.99J/g^oC\times (180-100)^oC]$

$\Delta H=1.16\times 10^6J$

Therefore, the amount of heat required is, $1.16\times 10^6J$

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3 years ago