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3 years ago

The concentration of hydronium ions is greater than 1x10-7 for basic solutions

1 answer:

8 0

FALSE

Concentration (M)= mol of substance/ given liter

OH- is hydroxide, a BASE
> proton acceptor
> produce hydroxide ions when dissolved in solution
H3O+ is hydronium, an ACID
> proton (H+) donor
> produce hydrogen ions when dissolved in solution

For more BASIC solutions:
concentration of BASE> con. of ACID
con. OH- > con. H3O+

The concentration of H3O+ must be less than or SMALLER than 1x10^-7 (so 1x10^-8, 1x10^-9 etc.) whereas the concentration of the base, OH-, must be LARGER than 1x10^-7. So there is a higher concentration of OH-, or more mol OH per L of solution.

pH measure the concentration of H+ or H3O+ ions in solution. It is -log[H+] ... so if the concentration was lower than 1x10^-7 mol/ L
-log(1x10^-8)= pH= 8 which is Basic (x>7)

Remember, for the pH scale,
(0-7)= acidic
7= neutral
(7-14)= basic

Conversely, for acidic solutions, the concentration of hydronium would increase

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Explanation:

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Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.970 atm when 25.5 g of glucose is consumed in this re

katrin2010 [14]

I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) ------> 6CO2(g) + 6H2O(l)

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3 years ago

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Flauer [41]

Answer:

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3 years ago

How many neutrons are present in an atom with atomic number 9 and mass number 19?

Kaylis [27]

Answer:

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5 0

3 years ago

Read 2 more answers

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago