What is the Sun's declination on the fall equinox? <br> a.0° <br> b.30° N <br> c.60° S <br> d.90° N

How accurate are heist movies where the robbers throw a bunch of gold bars in a bag and carry it out of the bank? Very, because

Oduvanchick [21]

Answer:

Not very, because gold is too heavy to carry many bars at once.

Explanation:

Such movies are far from being accurate.

<em>A standard gold bar weighs about 12.4 kilograms, an equivalence of approximately 27 pounds or 437 ounces, in real life. Hence, a bunch of gold bars would weigh far more than what one person can just lift up, throw, or carry as if it is a bunch of cotton wool. </em>

The correct answer is, therefore, that <u>such movies are far from reality because gold is too heavy to carry many bars at once</u>.

4 0

3 years ago

NH3 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Stels [109]

This question will be answered in parts:

<h3><u>1. Lewis Structure</u></h3>

There are three N-H bonds and one lone pair on the nitrogen atom in the Lewis structure of NH3. On hydrogen atoms, there are no lone pairs that can only hold two electrons.

The lewis structure is as shown in the diagram.

<h3><u>2. Perspective Drawing</u></h3>

A molecule is shown from a viewpoint with its atoms' bonds pointing either in your direction (bolded wedge) or away from you (hash wedge).

The perspective drawing of NH3 is given here

<h3><u>3.Number of atoms bonded to central atom </u></h3>

Nitrogen is the central atom in the Lewis structure of NH3, which also contains one lone pair and is bound to the three hydrogen atoms.

<h3><u>4.Number of non-bonding electron pairs on the central atom</u></h3>

Three bond pairs and two lone pairings exist. This is due to the fact that nitrogen contains five electrons in its outer shell, three of which are bound to hydrogen atoms and two of which are free.

<h3><u>5. Electronic geometry:</u></h3>

The electronic geometry of nitrogen is based on a tetrahedral arrangement of electron pairs, and ammonia likewise possesses four electron pairs. There is only one lone pair because there are only three connected groupings. But since the lone pairs are "invisible," the ammonia has a pyramidal form.

<h3><u>6. Molecular geometry with ideal bond angles</u></h3>

The molecular geometry of ammonia it has a trigonal pyramidal or distorted tetrahedral structure.

The bond angle in ammonia is less than the standard 109.5⁰. The bond angle is 107⁰

<h3><u>7.Hybridization of central atom</u></h3>

The core nitrogen (N) atom in the NH3 molecule has a steric number of (3 + 1) = 4, which causes Sp3 hybridization.

<h3><u>8. Polarity:</u></h3>

Because of its asymmetrical form, a trigonal pyramidal structure, and the different electronegativities of N(3.04) and H(2.2), the NH3 (ammonia) molecule is polar in nature .

To know more about ammonia, you can refer to:

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3 0

2 years ago

Which of the following ions is formed when a base is dissolved in a solution?

Aneli [31]

Answer:

$\huge\boxed{\sf OH-}$

Explanation:

<u>According to Arrhenius concept of acid and base:</u>

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

<u>For example:</u>

NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)

$\rule[225]{225}{2}$

Hope this helped!

3 0

3 years ago

(a.) a 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g).

Natali [406]

The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

Moles of CO2 formed can be calculated as

= Mass of CO2 / Molar mass of CO2

= 1.9061 / 44 = 0.0433 moles

<h3>Calculation of no. of moles of carbon</h3>

Now, moles of C which is present in one mole of CO2 = 1 mole

Moles of C in 0.0433 moles of CO2 = 0.0433 moles

As we know that, molar mass of C = 12 g / mol

Mass of C in 0.7549 g of given sample can be calculated as

= 0.0433 × 12 =0.5196 g

Mass of H2O formed = 0.3370 g

Similarly, Molar Mass of H2O = 18 g / mol

Moles of H2O = 0.3370 / 18 = 0.0187 moles

Moles of H present in 1 mole of H2O = 2 moles

Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles

Molar mass of H = 1 g / mol

Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g

Mass of O in 0.7549 g sample can be calculated as

= 0.7549 – [(Mass of C ) + (Mass of H) ]

= 0.7549 – [ (0.5196) + (0.0374) ]

= 0.1979 g

Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

learn more about Moles:

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DISCLAIMER: THE above question is incomplete. Complete question is given below:

A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass of C, H and O in the given sample.

4 0

2 years ago

Plz help me well mark brainliest if correct!!..

bagirrra123 [75]

Answer:

B

Explanation:

Nobody buliding a home in a rain forest unless they want to die and the rest make no sense

8 0

3 years ago

What is the Sun's declination on the fall equinox? a.0° b.30° N c.60° S d.90° N