1. A 205 g sample of H20 at 25.0 °C is heated until the temperature reaches 75.0 °C. How many

What is the volume, in liters, of 2.00 moles of hydrogen at STP?

Ber [7]

The volume of one mole of any gas at STP is 22.4 L. So, at STP, the volume of 2.00 moles of hydrogen gas would be (22.4 L/mol)(2 mol H2) = 44.8 L.

8 0

2 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

2 years ago

What geometry does VSEPR predict for the central atom in PF5

ddd [48]

so you can see those fluorine atoms have really spread out around the central phosphorus atom. this gives us a trigonal bi-pyramidal molecular geometry for pf5.

8 0

2 years ago

: calculate the mass of the solute in 1.500 l of 0.30 m glucose, c6h12o6, used for intravenous injection

vfiekz [6]

Answer is: the mass of the glucose is 81,07 grams.
c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.

8 0

3 years ago

How could you increase the atomic number?

TEA [102]

The increase in the number of atoms allows the strong positive charge of the nucleus to increase. Thus, due to the number of positive protons increasing in the nucleus, the positive charge increases. On the negatively charged electron cloud, the high positive charge of the nucleus has a strong tug.

8 0

2 years ago

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