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3 years ago

What kind of solid is crystalline boron

Chemistry

1 answer:

Ulleksa [173]3 years ago

7 0

Answer:

It is a network solid, a lattice of many covalent bonds (like diamond, except that it is black rather than transparent).

Explanation:

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One gram of an acid, HA of unknown molecular mass is dissolved in water and 22.72 cm3 of 0.60 M sodium hydroxide solution added

FinnZ [79.3K]

Chemistry

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3 years ago

A solution contains 1 LaTeX: \times\:×10−4 M OH– ions. Calculate the solution pH value, and determine if the solution is acidic,

docker41 [41]

Answer:

pH = 10

The solution is basic.

Explanation:

A solution contains 1 × 10⁻⁴ M OH⁻ ions. First, we will calculate the pOH.

pOH = -log [OH⁻]

pOH = -log 1 × 10⁻⁴

pOH = 4

We can find the pH of the solution using the following expression.

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 4 = 10

Since the pH > 7, the solution is basic.

5 0

4 years ago

A neutral atom has 37 electrons. How many protons does the atom have?

vlabodo [156]

Answer:

Explanation:

because when an element is neutral it contains thesame amount of protons and electrons.

3 0

3 years ago

If the temperature of air in gabor's lungs is 37∘c (98.6∘f), and the volume is 6l, how many moles of air n must be released by t

umka21 [38]

Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.

T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.

5 0

4 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago