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3 years ago

Which phrase describes an atom

Chemistry

1 answer:

pochemuha3 years ago

8 0

<span>A negatively charged electron cloud surrounding a positively charged nucleus. </span>

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Jade wants to measure the thickness of a copper wire. She wound the copper wire 30 times around a pencil and used a ruler to mea

Gala2k [10]

Answer: 0.05

Explanation:Divide the length (1.5 cm) by the number of turns (30)

3 0

3 years ago

Read 2 more answers

D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match th

Aleks04 [339]

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

6 0

3 years ago

What is an atomic weight?

S_A_V [24]

The average weight of an atom of an element, formerly based on the weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the weight of an oxygen atom, but after 1961 based on 1/12 the weight of the carbon-12 atom.

3 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

What is the kinetic energy of a 1-kilogram ball is thrown into the air with ab ititial velocity of 30m/s?

mariarad [96]

KE=1/2*mass*velocity^2
So u do 1/2 * 1 * 30^2
1/2 * 1 * 900
= 450kgm/s

P.s. I'm not sure if I would have to convert kg to g.
Anyways hope this helped

4 0

3 years ago