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4 years ago

Write a brief account of Ernest Rutherford’s personal and professional life in no more then 150 words

1 answer:

3 0

<span><span> i think the answer is that >>>>>Ernest Rutherford was born in rural Spring Grove, on the South Island of New Zealand on August 30, 1871. He was the fourth of 12 children, and the second son. His father, James, had little education and struggled to support the large family on a flax-miller’s income. Ernest’s mother, Martha, worked as a schoolteacher. She believed that knowledge was power, and placed a strong emphasis on her children’s education.As a child, Ernest, whose family called him “Ern,” spent most of his time after school milking cows and helping with other chores on the family farm. Weekends were spent swimming in the creek with his brothers. Since money was tight, Rutherford found inventive ways of overcoming his family’s financial challenges, including birds-nesting to earn funds for his kite-flying supplies. “We haven’t the money, so we’ve got to think,” was Rutherford’s motto at the time.<span><span /></span>
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Magnesium bromide is a binary ionic compound. From its formula, MgBr2, how do you know that

Scilla [17]

You that Mg is the metal if in the periodic table it lands between group 1-12. So yeas Mg is a metal because it lands in group 2.

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4 years ago

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Please help will reward brainliest

dimaraw [331]

The answer is most definitely “A”

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3 years ago

On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sk

timurjin [86]

Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

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3 years ago

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Draw the Lewis structure for NF3. What orbitals on N and F overlap to form bonds between these elements

jonny [76]

Is there a picture ????????????

4 0

3 years ago

Need help asap with this chemistry if someone could help me

Burka [1]

Answer:

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

Structure Number Two would likely be the most stable structure.

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

The N atom is the one that is "likely" to be attracted to an anion. See explanation.

Explanation:

When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:

$\text{Formal Charge} \\ = \text{Number of Valence Electrons in Element} \\ \phantom{=}-\text{Number of Chemical Bonds} \\\phantom{=} - \text{Number of nonbonding Lone Pair Electrons}$ .

For example, for the N atom in structure one of the first question,

N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
This N atom is connected to only 1 chemical bond.
There are three pairs, or 6 electrons that aren't in a chemical bond.

The formal charge of this N atom will be $5 - 1 - 6 = -2$ .

Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.

Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.

Similarly,

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.

Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.

5 0

3 years ago