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grandymaker [24]
2 years ago
6

A pair of molecular orbitals is formed by.

Chemistry
1 answer:
miskamm [114]2 years ago
3 0
“Bonding molecular orbitals are formed by... in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule.”
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2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?
sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

Q = c \times m \times \Delta T

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

m = \frac{Q}{c \times \Delta T  }  = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C)  } = 672 g

The mass of water that can be warmed under these conditions is 672 grams.

5 0
3 years ago
What is diffusion ,?? ​
UNO [17]
Diffusion is a process that results from the random motion of molecules and results in a net movement of matter from a high-concentration region to a low-concentration zone.
4 0
2 years ago
Read 2 more answers
Lol someone help me with this
lozanna [386]

Answer:

Its saying how many electronics are in your house.

Explanation:

Such as a tablet or a tv.etc

7 0
2 years ago
A 1.757-g sample of a / alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of
kondor19780726 [428]

Answer:

78.14% Pb²⁺ and 21.86% of Cd²⁺

Explanation:

The first titration involves the reaction of both Pb²⁺ and Cd²⁺

In the second titration, as the buffer is HCN/NaCN, the Cd²⁺ precipitates as Cd(CN)₂ and the only ion that reacts is Pb²⁺

In the first titration:

<em>Moles EDTA = Moles Pb²⁺ and Cd²⁺:</em>

28.89mL = 0.02889L * (0.06950moles / L) = 2.008x10⁻³ moles in the aliquot. In the sample:

2.008x10⁻³ moles * (250.0mL / 50.0mL) =

0.01004 moles = Pb²⁺ + Cd²⁺ <em>(1)</em>

In the second titration:

19.07mL = 0.01907L * (0.06950mol / L) = 1.325x10⁻³ moles Pb²⁺ in the aliquot. In the sample:

1.325x10⁻³ moles Pb²⁺ * (250.0mL / 50.0mL) =

6.626x10⁻³ moles Pb²⁺

That means the moles of Cd²⁺ are:

0.01004 moles = Cd²⁺ + 6.626x10⁻³ moles Cd²⁺

3.413x10⁻³ moles Cd²⁺

The mass of each ion is:

<em>Cd²⁺ -Molar mass: 112.411g/mol-:</em>

3.413x10⁻³ moles Cd²⁺ * (112.411g / mol) =

0.384g of Cd²⁺

<em>Pb²⁺ -Molar mass: 207.2g/mol-:</em>

6.626x10⁻³ moles Pb²⁺ * (207.2g / mol) =

1.373g of Pb²⁺

The percent mass of each ion is:

1.373g Pb²⁺ / 1.757g = 78.14% Pb²⁺

And:

0.384g of Cd²⁺ / 1.757g * 100 = 21.86% of Cd²⁺

4 0
3 years ago
Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutraliz
madreJ [45]

Answer:

0.197 M

Explanation:

The reaction equation is:

H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M

4 0
2 years ago
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