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fomenos
3 years ago
6

How many atoms are in 3.5 moles of arsenic atoms?

Chemistry
2 answers:
crimeas [40]3 years ago
8 0

Answer:

2.1 x 1024 atoms

Explanation:

This is the right, trust me :)

solmaris [256]3 years ago
5 0

Answer:

2.107×10²⁴

Explanation:

no of moles=no of particles/Avogadro's number

3.5=X/6.02×10²³

by cross multiplication

3.5×6.02×10²³

=2.107 ×10²⁴

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What is the number of valence electron of (Na). ?
sammy [17]
There is 11 valence electrons in sodium (Na)
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A science fair volcano bubbles and fizzes. what is taking place?
vaieri [72.5K]
Answer is B. gas formation
7 0
3 years ago
The period of a ripple on a lake is half a second. what is the frequency of the wave
Nataly_w [17]

Answer:

2 Hertz

Explanation:

<em>The frequency would be 2 Hertz.</em>

<u>The frequency of a wave is defined as the rate at which the particles of a medium vibrates when the wave is passed through it while the period of a wave is the time it takes the particles to make a complete cycle of vibration.</u>

The frequency of a wave is inversely related to its period and is defined by the following equation:

f = 1/t, where f is the frequency (in hertz) and t is the period (in seconds).

Hence, if the period of a ripple is 1/2 or 0.5 seconds, the frequency becomes;

f = 1/0.5 = 2 Hertz

7 0
3 years ago
South Florida is surrounded by salt water. The need for fresh
Lady bird [3.3K]

Answer:

Aquifer

Explanation:

Instead of snowcapped mountains that store water in advance of warmer temperatures, most of our drinking water comes from underground "mountains" of porous materials called aquifers which are replenished by rain. The Biscayne Aquifer is South Florida's lower east coast's primary source of fresh water.

3 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
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