The water cycle, photosynthesis, nutrient cycling, and soil formation are examples of:

Determine the number of valence electrons for the following: [kr] 5s2 4d6

wlad13 [49]

Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).

Explanation:

This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7

Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4

3 0

3 years ago

At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar

Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr

The partial pressure of acetic acid is 73.5 torr

4 0

3 years ago

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

(b) Moles of $H_2SO_4$ can be calculated as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

Moles of $H_2SO_4$ = 0.00145 moles

From the reaction,

1 mole of $H_2SO_4$ react with 2 moles of NaOH

0.00145 mole of $H_2SO_4$ react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M

7 0

3 years ago

The thin muscular wall that separates the right side of the heart from the left side is called the

Lina20 [59]

The upper chambers are called the left and right atria, and the lower chambers are called the left and right ventricles. A wall of muscle called the septum separates the left and right atria and the left and right ventricles

8 0

3 years ago

Is PH3 polar or nonpolar

Andru [333]

Answer:

polar

Explanation:

8 0

3 years ago

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