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3 years ago

Which conditions are necessary for a lunar eclipse to occur?

2 answers:

riadik2000 [5.3K]3 years ago

8 0

It'd be either A or D.
When a lunar eclipse begins, the moon goes in front of the sun and blocks out the sun light, making it completely dark.

vekshin13 years ago

4 0

The answer is B) the moon is in Earth's shadow

You might be interested in

In the chemical reaction:

velikii [3]

Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 1 mole
AgNO₃: 2 moles
Ag: 2 moles
Cu(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}$

<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

<h3>Summary</h3>

In summary, AgNO₃ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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8 0

1 year ago

Which of the following statements about tornadoes is false?

loris [4]

The answer is A because the time wouldn't affect the weather that much depending on the circumstances.

4 0

3 years ago

Read 2 more answers

How can a short-term environmental change, such as a drought or a flood, affect organisms?

viktelen [127]

Answer:

it can affect things by drastically chagning the way that organisms opareate such as the eco systems, the health of the land the flood or drought is on and etc.

Explanation:

hope this helps!

5 0

2 years ago

Which of earths spheres contains most of its mass?

zysi [14]

The geosphere is about 99.94% of Earth's mass, so C is the answer.

8 0

3 years ago

Read 2 more answers

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago