The average atomic mass of the imaginary element : 47.255 amu
<h3>Further explanation </h3>
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% ..
isotope E-47 47.011 amu, 87.34%
isotope E-48 48.008 amu, 6.895
isotope E-49 50.009 amu, 5.77%
The average atomic mass :
Answer:
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ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
See explanation
Explanation:
The periodic table shows the atomic number and mass number of each element.
We know that the atomic number shows;
- The number of protons in the nucleus of the atom
- The number of electrons in the neutral atom of the element.
So we obtain the number of protons and electrons by looking at the atomic number shown in the periodic table.
We also know that;
Mass number = Number of protons + number of neutrons
Since number of protons = atomic number of the atom
Number of neutrons = Mass number - atomic number
Hence we obtain the number of protons by subtracting the atomic number from the mass number given in the periodic table.
Answer:
3.676 L.
Explanation:
We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
Knowing that:
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
Answer:
78
Explanation:
That's my score when I did a 3-minute Step test
