What is the Kelvin temperature for 30°C?

What is force?.......<br><br>

snow_tiger [21]

<em>Force is a push or pull which changes or tend to change the position of a body...</em>

4 0

2 years ago

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An ionic bond is between what kind of charged particles?

Lisa [10]

Answer: An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.

Explanation:

3 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

2 years ago

BALANCE the following equations then write the NET IONIC EQUATION for each one: ____Ni(NO 3 ) 2(aq) + ____NaOH (aq) →

Karo-lina-s [1.5K]

Answer:

Ni(NO3)2 + 2NaOH → Ni(OH)2 + 2NaNO3

NET IONIC EQUATION :

Ni +2(aq) + 2(NO3) -1(aq) + 2Na +1(aq) + 2OH -1(aq)

→ Ni(OH)2 (s) + 2Na +1(aq) + 2(NO3) -1(aq)

sorry there is no space to write the reaction in one line...

8 0

3 years ago

HELP I WILL mark brainliest

nadya68 [22]

Answer:

1 Hour...

Explanation:

If It’s 960 Km/h, then it will take 1 hour to fly there.

4 0

2 years ago