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3 years ago

11. (03.03 MC)

Chemistry

1 answer:

Stells [14]3 years ago

6 0

Answer:

A) An oxygen atom shares an electron pair with each H atom.

Explanation:

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In a metabolic pathway, succinate dehydrogenase catalyzes the conversion of succinate to fumarate. the reaction is inhibited by

Pani-rosa [81]

Competitive inhibitor

7 0

4 years ago

please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting react

VikaD [51]

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

Al₂O₃ + 3H₂O → 2Al(OH)₃

Number of moles;

Number of moles = $\frac{mass}{molar mass}$

molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole

number of moles = $\frac{3.87}{102}$ = 0.04mole

molar mass of H₂O = 2(1) + 16 = 18g/mole

number of moles = $\frac{5.67}{18}$ = 0.32mole

From the reaction equation;

1 mole of Al₂O₃ reacted with 3 moles of H₂O

0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant

6 0

3 years ago

The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ

vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

$q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m} =125.28W$

The heat flux can be defined as the amount of heat flow by unit of area.

Using the previous calculation, we can estimate the heat flux:

$heat \, flux=\frac{q}{A}=\frac{125.28 W}{9 m^{2} } =13.92 W/m^{2}$

It can also be calculated as:

$q/A=-k*\frac{\Delta T}{\Delta X}$

The thermal resistance can be expressed as

$\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}$

For the 3m x 3m sheet, the thermal resistance is

$R_t = \frac{\Delta X}{k*A}=\frac{0.025m}{0.029W/mK*9m^{2}}=0.0958 \, K/W$

4 0

4 years ago

Pb has an atomic number of 82. It has an isotope Pb-207. The mass number is 207.

nikdorinn [45]

Answer:

Number of protons = 82

Explanation:

Given data:

Atomic number of Pb = 82

Mass number of Pb-207 = 207

Number of protons = ?

Solution:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

Atomic number of Pb = number of protons or electrons

82 = number of protons or electrons

Mass number = Number of protons + number of neutron

207 = 82 + number of neutron

Number of neutrons = 207 - 82

Number of neutrons = 125

5 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

4 years ago