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2 years ago

2) Examine the molecules below. Circle the molecules that can also be classified as compounds.

Chemistry

1 answer:

sergejj [24]2 years ago

4 0

Answer:

3rd

Explanation:

its the way they look

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Given that w for water is 2. 4×10−14 m^2 at 37 °C. Calculate the ph of a neutral aqueous solution at 37 °c, which is the normal

alexgriva [62]

The pH of a neutral aqueous solution at 37°C is 6.8.

Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:

Kw = [OH⁻] . [H₃O⁺]

Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.

<h3>Calculation</h3>

Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:

Kw =[OH⁻] . [H₃O⁺] = x²

x² = 2.4 × 10⁻¹⁴ M²

x = 1.5919 × 10⁻⁷ M

Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M

pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)

pH = 6.8

Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.

learn more about pH:

brainly.com/question/9529394

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3 0

2 years ago

The size of the chosen symbols on a graph may be used to indicate the:

Tresset [83]

Uncertainty of the data hope its help you

6 0

3 years ago

Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P(s) + 3 I2(g) → 2 PI

melamori03 [73]

Answer:

Percent yield of PI3 = 95.4%

Explanation:

This is the reaction:

2P (s) + 3I2 (g) > 2PI3 (g)

Let's determine the moles of iodine that has reacted.

58.6 g / 253.8 g/mol = 0.231 mol

Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.

3 moles of I2 react to make 2 moles of PI3

0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3

As we have produced 0.147 moles let's determine the percent yield.

(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%

5 0

3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago

In need of help very soon please.

NemiM [27]

Answer:

hello, i hope this helps.

Explanation:

1 - group

2 - period

3 - periodic table

4 - family

5 - octet rule

6 - valence electrons

6 0

3 years ago