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3 years ago

Ill give u brainliest plz help

1 answer:

6 0

Photosynthesis is when a plant takes in carbon dioxide produced by all breathing organisms and reintroduces oxygen into the atmosphere also creating sugar (glucose), it is so important for humans and animals because it provides the basic energy source for virtually all organisms

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Calculate the Ka for the following acid. Determine if it is a strong or weak acid. HClO2(aq) dissolves in aqueous solution to fo

Talja [164]

Answer:

The value of Ka $= 1.1*10^{-2}$

It is a weak acid

Explanation:

From the question we are told that

The concentration of $[HClO_2]=0.24M$

The concentration of $[H^+]=0.051M$

The concentration of $[ClO_2^-]=0.051M$

Generally the equation for the ionic dissociation of $HClO_2$ is

$HClO_2_(aq) -------> H^{+}_{(aq)} + ClO_2^{-}_{(aq)}$

The equilibrium constant is mathematically represented as

$Ka = \frac{concentration \ of \ product }{concentration \ of \ reactant }$

$= \frac{[H^+][ClO_2^-]}{[HClO_2]}$

Substituting values since all value of concentration are at equilibrium

$Ka = \frac{0.051 * 0.051}{0.24}$

$= 1.1*10^{-2}$

Since the value of is less than 1 it show that in water it dose not completely

disassociated so it an acid that is weak

3 0

3 years ago

HELP HELP HELP HELP!!!!!!

suter [353]

Answer:

I think it is 100 if I guess

8 0

2 years ago

Why does potassium permanganate need to be standardized right before a titration?.

Ierofanga [76]

To determine the strength of potassium permanganate with a standard solution of oxalic acid.

6 0

2 years ago

if the pressure of a gas at constant volume is 3.5 atm at 100°c, what will the pressure be if the tempature is changed to 250°c?

Degger [83]

Sorry don't know this one

6 0

3 years ago

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