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kumpel [21]
3 years ago
6

What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza

tion? What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutralization? 0.600 M 0.0300 M 0.750 M 0.150 M 7.50 M
Chemistry
1 answer:
puteri [66]3 years ago
5 0

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

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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
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