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pshichka [43]
3 years ago
5

Where are the protons, neutrons, and electrons located in the atom?

Chemistry
2 answers:
Aleksandr [31]3 years ago
6 0
Protons and neutrons are located in the nucleus of the atom. Electrons are located in the electron cloud outside/around the nucleus. Protons have a positive charge, neutrons have a neutral charge, or no charge, and electrons have a negative charge
Effectus [21]3 years ago
4 0
The Protons and Neutrons are both located in the nucleus. The protons have a positive charge, the neutrons have no charge, and electrons have a negative charge.
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Cells carry out which of the following functions in an organism?
mina [271]

Answer:

smaller organelles

4 0
3 years ago
How many moles are in 25.3 g of a sample of potassium nitrate (KNO3)?
aksik [14]

Answer:

25.3 g of KNO₃ contain 0.25 moles.

Explanation:

Given data:

Number of moles of KNO₃ = ?

Mass of KNO₃ = 25.3 g

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of KNO₃:

KNO₃ = 39 + 14+ 16×3

KNO₃ = 101 g/mol

Now we will put the values in formula.

Number of moles = 25.3 g/ 101 g/mol

Number of moles = 0.25 mol

Thus, 25.3 g of KNO₃ contain 0.25 moles.

8 0
3 years ago
Primary succession is most likely caused by?
disa [49]

Answer:

volcanic eruption.

mark my answer as brainlest......

3 0
3 years ago
How many moles of HCI would you have if you have 7 L of a 9M solution<br> of HCI? *
frutty [35]

Explanation:

1kg x 0,37=370gHCL

370g/36,5g/mol=10,137mol/kg

1l=1,185kg

10,137 x 1,185kg/l =12,0mol/l

(x/ml(12mol HCL)*12mol/L)/100ml=0,025mol/L

x=

5 0
3 years ago
Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
4 years ago
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