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2 years ago

11

What ion is at a higher concentration in the extracellular fluid than in the intracellular fluid?

1 answer:

Phoenix [80]2 years ago

8 0

There are two types of fluid in the body extracellular fluid and intracellular fluid (ECF and ICF), together they are account for total body water.
The Sodium (Na+) ion is at higher concentration in the extracellular fluid than in the intracellular fluid. The function of extracellular fluid is that it provide cells to watery environment so that they can easily live and perform their function.

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What is needed to burn the candle (reactant)?

Alexus [3.1K]

Answer:

wax, candlewick, and oxygen

Explanation:

The burning of the candle is both a physical as well as a chemical change. The reactants are the substances or the raw materials that are required for a reaction to the process. In the process of burning a candle, the reactants are the fuel which includes wax and wick, and oxygen which is found in the air. The products found at the end of the reaction are carbon dioxide and water vapor.

8 0

3 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$0.173\%=\frac{779 mg}{m}\times 100$

$m=\frac{779 mg}{0.173}\times 100= 450,289 mg$

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0

3 years ago

Determine the mass of 4.25x10^26 formula units of barium hydroxide

Andru [333]

$4.25x10x^{26}$

5 0

2 years ago

Hydrogen sulfide burns form sulfur dioxide:

Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $H_2S$

$\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles$

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ $\Delta H=-1036kJ$

According to stoichiometry :

2 moles of $H_2S$ on burning produces = 1036 kJ

Thus 0.78 moles of $H_2S$ on burning produces = $\frac{1036kJ}{2}\times 0.78=404.04$

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0

2 years ago

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

Lady_Fox [76]

Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0

2 years ago