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3 years ago

The function of a specific enzyme is most directly influenced by its

1 answer:

7 0

The function of a specific enzyme, a certain biological catalyst is most directly influenced by its shape and structure. The shape of the enzyme, primarily determined by the composition of amino acids that constitute or make up the primary structure of it. Will vary in terms of the interactions between these amino acids, and various parts within it. Observed in later levels of protein organization.

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Could someone please help!

bezimeni [28]

Im pretty sure that the current will increase so the answer is A . Hope that helps

8 0

4 years ago

I really need help with b. Please help

fiasKO [112]

Answer:

Hi Im an online tutor and i can assist you with all your assignments. We have experts in all fields. check out our website https://toplivewriters.com/

Explanation:

8 0

2 years ago

1. The electron arrangement ions X and Y2 are 2, 8, and 2, 8, 8 respectively.

svetoff [14.1K]

The given question is incomplete. the complete question is:

The electron arrangement ions $X^{3+}$ and $Y^{2-}$ are 2, 8, and 2, 8, 8 respectively.

Write the electronic arrangement of the elements X and Y.

Write the formula of the compound that would be formed between X and Y.

Answer: 1. The electronic arrangement for X and Y are 2,8, 3 and 2,8,5 respectively.

The formula of the compound formed is $X_2Y_3$

Explanation:

Electronic configuration is the arangement of electrons in an atom in order of increasing energies.

Cations are formed when electrons are lost by atoms and anions are formed when electrons are gained by atoms.

$X^{3+}:2,8$

$X:2,8,3$

$Y^{3-}:2,8,8$

$Y:2,8,5$

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

Here element X is having an oxidation state of +3 called as $X^{3+}$ cation and $Y^{2-}$ is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $X_2Y_3$

3 0

3 years ago

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1

nordsb [41]

Answer: The enthalpy change of the unknown solid is 3.824 kJ/mol

Explanation:

To calculate the mass of solution , we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.20 g/mL

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$1.20g/mL=\frac{\text{Mass of solution}}{150.0mL}\\\\\text{Mass of solution}=(1.20g/mL\times 150.0mL)=180g$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T$

q = heat absorbed or released

m = mass of solution = 180 g

c = specific heat capacity = 4.18 J/g°C

$\Delta T$ = change in temperature = 16.01°C

Putting values in above equation:

$Q=180\times 4.18\times 16.01=12045.9J=12.046kJ$

To calculate the enthalpy change of the reaction, we use the equation

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 12.046 kJ

n = number of moles of solid = 3.15 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{12.046kJ}{3.15mol}=3.824kJ/mol$

Hence, the enthalpy change of the unknown solid is 3.824 kJ/mol

5 0

3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

4 0

4 years ago