Which of the following is the best example of a "Scientific Question"?

Consider a neutrally charged atom that has an atomic mass number of 80, which includes 35 neutrons. how many electrons does this

ivanzaharov [21]

Atomic mass number is the number of protons and neutrons. Subtract 80-35=45 is the number of protons. Because the atom is neutrally charged, the number of protons must equal the number of electrons, so there are 45 electrons.

5 0

3 years ago

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An aqueous solution of Calcium Chloride reacts with an aqueous solution

Tems11 [23]

Answer:

the products formed are : -

1. CaCO3 - Calcium Carbonate

2. NaCl - Sodium Chloride

Explanation:

Calcium chloride reacts with Sodium carbonate to form Calcium carbonate and Sodium chloride. this reaction is a double displacement reaction.

here's the balanced chemical equation for the above reaction : -

CaCl2 + Na2CO3 =》CaCO3 + 2 NaCl

7 0

2 years ago

Can you please help me it’s not a hard question

aliya0001 [1]

The answer for acceleration is m/s^2

5 0

2 years ago

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

wolverine [178]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2) $5.6\cdot 10^{13}$ electrons per person

Explanation:

1)

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

$v_d=\frac{I}{neA}$

where

I is the current

n is the concentration of free electrons

$e=1.6\cdot 10^{-19}C$ is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

$A=\pi r^2$

where r is the radius of the wire. So the equation becomes

$v_d=\frac{I}{ne\pi r^2}$

In this problem, we have:

I = 8.0 A is the current

$8.5\cdot 10^{28} m^{-3}$ is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

Therefore, the drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

2)

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Therefore, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , so the total number of electrons in this cord is

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

In total, the Earth population consists of 8 billion people, which is

$N'=8\cdot 10^9$

Therefore, the number of electrons that each person would get is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

3 years ago

Polysaccharides are simple sugars

choli [55]

No polysaccharides are complex forms of carbohydrate of long sugar molecules

7 0

2 years ago