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2 years ago

Which chemical formula represents both an element and a molecule?

Chemistry

1 answer:

victus00 [196]2 years ago

5 0

Answer:

Explanation:

element - a basic substance that can't be simplified (hydrogen, oxygen, gold, etc...) molecule - two or more atoms that are chemically joined together (H2, O2, H2O, C6H12O6, etc...) compound - a substance that contains more than one element (H2O, C6H12O6, etc...)

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Which part of the cell is affected by the movement of molecules through diffusion, osmosis, and active transport? A.Cell membran

Serjik [45]

Answer:

membrana celular

Explanation:

El transporte celular es el intercambio de sustancias a través de la membrana plasmática, que es una membrana semipermeable.1

El transporte es importante para la célula porque le permite expulsar de su interior los desechos del metabolismo, también el movimiento de sustancias que sintetiza como hormonas. Además, es la forma en que adquiere nutrientes mediante procesos de incorporación a la célula de nutrientes disueltos en el agua. Las vías de transporte a través de la membrana celular y los mecanismos básicos para las moléculas de pequeños tamaños son:

Índice

1 Transporte pasivo

1.1 Ósmosis

1.1.1 Ósmosis en una célula animal

1.1.2 Ósmosis en una célula vegetal

1.2 Difusión facilitada

2 Transporte activo

2.1 Transporte activo primario: Bomba de sodio y potasio o Bomba Na+/K+

2.2 Transporte activo secundario o cotransporte

3 Transporte en masa

3.1 Endocitosis

3.2 Exocitosis

4 Véase también

5 Referencias

6 Enlaces externos

6 0

2 years ago

We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of

Arte-miy333 [17]

Answer:

A. pH using molar concentrations = 2.56

B. pH using activities = 2.46

C. pH of mixture = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

HA + H₂O ⇌ A⁻ + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}$

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

$\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}$

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is

$I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041$

(iii) Calculate the activity coefficients

$\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79$

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH

$\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\$

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

$I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10$

(ii) Calculate the activity coefficients

$\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69$

(iii) Calculate the initial activity of A⁻:

a = γc = 0.69 × 0.05= 0.034

(iv) Calculate the pH

$\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}$

3 0

3 years ago

If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

bulgar [2K]

<u>Answer:</u> 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

<u>Explanation:</u>

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = $\frac{1}{2}\times 6.5=3.25moles$ of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = $\frac{2}{2}\times 6.5=6.5moles$ of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

7 0

3 years ago

Which of these reactions are redox reactions and why?

IceJOKER [234]

It will be number 1.
NO loses oxygen ie its being reduced to N2
NH3 gains oxygen atom ie its being oxidised to H2O

7 0

3 years ago

What is the concentration of hydronium ion in a 0.121 M HCl solution?

kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0

3 years ago