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2 years ago

PLEASE HELP!!!!!!!!!!!!!

Chemistry

1 answer:

monitta2 years ago

8 0

Answer:

• The closing of the frontier

Explanation:

•• The closing of the frontier• The closing of the frontier• The closing of the frontier• The closing of the frontier• The closing of the frontier

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For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

3 0

3 years ago

What is the balanced equation to, CH4+O2=C02+H2O ?

Alecsey [184]

CH4 + 2O2 --> CO2 + 2H2O

6 0

3 years ago

Read 2 more answers

15 mL of a 0.16 M solution of glucose was diluted until the total volume reached 95 mL. What is the concentration of new solutio

Nana76 [90]

Answer:

the concentration of new solution = 0.025 M

Explanation:

Given that :

The initial volume V_1 = 15 mL

Initial concentration M_1 = 0.16 M

Final volume V_2 = 95 mL

Final concentration M_2 = ???

We know that M_1 × V_1 = M_2V_2

Making M_2 the subject of the formulae; we have :

M_2 = (M_1 × V_1)/V_2

M_2 = (0.16 × 15)/ 95

M_2 = 0.025 M

Thus, the concentration of new solution = 0.025 M

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3 years ago

What does it mean to transcribe DNA to RNA

aksik [14]

Answer:

EIt involves copying a gene's DNA sequence to make an RNA molecule. Transcription is performed by enzymes called RNA polymerases, which link nucleotides to form an RNA strand (using a DNA strand as a template). ... Transcription is controlled separately for each gene in your genome.xplanation:

8 0

3 years ago