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Akimi4 [234]
2 years ago
14

Adding one proton to the nucleus of an atom

Chemistry
1 answer:
Sonbull [250]2 years ago
7 0
Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.
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El amoniaco y el fluor reaccionan para formar tetrafluoruro de dinitrogeno y fluoruro de hidrogeno. segun la reaccion: NH3 + F2
White raven [17]

Answer:

son 12.6 gramos de HF

Explanation:

Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos

5 0
2 years ago
Elena has two magnets. She puts one under a piece of paper. She holds the other one above it. The magnets attract each other. Th
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A) Magnets can attract through solid materials.
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2 years ago
In two or more complete sentences describe all of the van der Waals forces that exist between molecules of sulfur
zavuch27 [327]

Answer:

Dipole-Dipole attraction

Explanation:

Dipole-dipole attraction is a type of vander waals forces found in the molecules of sulfur dioxide.

Vander waals forces are weak attractions joining non-polar and polar molecules together. They are of two types:

  • London dispersion forces which are weak attractions found between non-polar molecules.
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3 0
2 years ago
Read 2 more answers
14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its
masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

7 0
2 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =
IgorC [24]
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
  = 2.16 x10^2
8 0
2 years ago
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