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koban [17]
2 years ago
9

An atom contains 5 protons, 6 neutrons, and 6 electrons. What is the mass number of this atom? *

Chemistry
1 answer:
Tju [1.3M]2 years ago
6 0

Answer:

here you go

Explanation:

Atomic Number (Z) = Mass Number (A) - Number of Neutrons An atom has 5 protons, 5 electrons and 6 neutrons The atomic number = number of protons = number of electrons = 5 The mass number = 5 protons + 6 neutrons = 11

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How many grams<br>are in 3.0 moles of KCI<br>(potassium chloride)?​
Ganezh [65]

Answer: I thinc 223.6539

Explanation:

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1. Suppose during volleyball practice, you lost 2.0 lbs of water due to sweating. If all of this
Lynna [10]

Answer:

E=2052.8 kJ

Explanation:

The energy absorbed by the water is the energy it requires to evaporate. So:

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The moles of water:

n_w=m_w *\frac{1000g}{2.2lbs}*\frac{1}{M}

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n_w=2 lbs *\frac{1000g}{2.2lbs}*\frac{1}{18g/mol}

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3 0
3 years ago
It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J 2.80×107 J of
Alborosie

Answer:

The question is incorrect and incomplete. Here's the correct question:

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J  of energy when burned. To illustrate this difficulty,a) calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L  of crude oil, if the water has its temperature raised from 23.5 °C to 100 °C , it boils, and the resulting steam is raised to 315 °C. b)Discuss additional complications caused by the fact that crude oil has less density than water.

Explanation:

Q= mc ΔT

Q= heat energy

m is mass

ΔT is change in temperature and c is specific heat capacity

calculating heat for latent heat of vaporisation

Q= ml where l is latent heat of vaporisation

a) Total heat energy used= heat required to raise temperature from 23.5 °C to 100 °C, heat required to boil water and heat required to further raise temperature from 100 °C to 315°C

Q = mc ΔT₁ + mL + mc ΔT₂

Q = m(c ΔT₁ + L + c ΔT₂)

m= Q÷(c ΔT₁ + L + c ΔT₂)

Q= 2.8 X 10⁷ J

c=4186J/kg°C

L=2256 x 10³J/kg

ΔT₁=76.5°C(100°C-23.5°C)

ΔT₂= 215°C(315°C-100°C)

(c ΔT₁ + L + c ΔT₂)= 4186J/kg°C *76.5°C + 2256 x 10³J/kg + 4186J/kg°C*215°C =3476219J/Kg

m= 2.8 x 10⁷J ÷3476219J/Kg

m =80.54 Kg

volume = mass÷ density

=80.54kg ÷ 10³kg/m³( density of water)

=0.0854m³

0.001m³ = 1 lL0.08054m³= 0.08054m³ /0.001m³= 80.54L

VOLUME is 80.54litres

b) since the density of crude is less than the density of water,and 80L of additional water is added, it'll make the crude to float on water thus inhibiting the extinguishing process

4 0
3 years ago
Elements are distinguished from each other by the number of protons present in their nuclei??
Kisachek [45]

False....................

7 0
2 years ago
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