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2 years ago

What pH must organisms maintain

Chemistry

2 answers:

vazorg [7]2 years ago

7 0

It's different for each organ. So in the stomach its gotta be acidic and in your blood its gotta be neutral

MA_775_DIABLO [31]2 years ago

5 0

6.5-7.5 i believe was right

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A solution is prepared by dissolving 15.0 g of NH3 in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The

iVinArrow [24]

Answer:

[NH₃] → 3.24 M

Explanation:

Our solute: Ammonia

Our solvent: Water

Solution's mass = Mass of solute + Mass of solvent

Solution's mass = 15 g + 250 g = 265g

We use density to determine, the volume.

D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.

We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L

To determine molarity we need the moles of solute in 1 L of solution.

Moles of solute are: 15g / 17g/mol = 0.882 moles

[NH₃] = 0.882mol /0.27207 L → 3.24 M

3 0

3 years ago

A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma

klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, $m_1$ = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, $m_2$ = $(M-m_1)=(39.664-21.577)g=18.087g$

To calculate density of a substance, we use the equation:

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

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

$\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL$

Hence, the density of NaCl solution is 3.930 g/mL

7 0

2 years ago

Solid solutions that are mixtures of elements with metallic properties are known as_____

marshall27 [118]

Alloys. hope this helps.

4 0

2 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.