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Paraphin [41]
3 years ago
5

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

atom have?
Chemistry
2 answers:
ivanzaharov [21]3 years ago
5 0
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.
ycow [4]3 years ago
3 0

Answer:

45 electrons.

Explanation:

mass number = 80

no. of neutrons = 35

Given that it is a neutrally charged atom,

no. of electrons = mass no. - no. of neutrons

                          =    80 - 35

                          =      45 electrons.

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A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0
3 years ago
Identify the element in C6H12O6 that allows it to be classified as an organic compound.
LenaWriter [7]

Answer:

i dont know lol

Explanation:

8 0
3 years ago
What is the volume of 0.20 moles of helium at STP?
stealth61 [152]

The volume of 0.20 moles of helium at STP is 4.5 liters.

Explanation:

Given:

Number of moles  = 0.20 moles

To Find:

The volume of Helium at STP =?

Solution:

According to ideal gas law

PV = nRT

where

P is pressure,  

V  is volume,  

n  is the number of moles  

R  is the gas constant, and  

T  is temperature in Kelvin.

The question already gives us the values for  p  and  T ,because helium is at STP. This means that temperature is  273.15 K  and pressure is  1 atm .

We also already know the gas constant. In our case, we'll use the value of  

0.08206 L atm/K mol  since these units fit the units of our given values the best

On substituting these values we get

1 atm \times  V = 0.20 moles  \times  0.08206 L atm/K mol

V = \frac{0.20 moles  \times  0.08206 L atm/K mol }{ 1 atm}

V = \frac{4.482}{1}

V =  4.5 Liters

7 0
3 years ago
Please use the below balanced equation to answer this question.    2H2(g)  +  O2(g)  ->   2H2O(l) How many grams of water wil
Blizzard [7]
At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L 
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D

4 0
3 years ago
Which isotope is used to treat cancer?<br> (1) C-14 (3) Co-60<br> (2) U-238 (4) Pb-206
lions [1.4K]
I believe doctors use (3) Co-60 to treat ILWs. Uranium and Carbon are not used, and a different isotope of lead is used to be attached to monoclonal antibodies.
8 0
3 years ago
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