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3 years ago

How many moles of KOH are there in 27.5 mL of 0.250 M KOH?

Chemistry

1 answer:

zaharov [31]3 years ago

7 0

Answer:

6.88 × 10^-3mol

Explanation:

Molarity = number of moles (n) ÷ volume (V)

According to this question, there in 27.5 mL of 0.250 M KOH, the number of moles of KOH can, therefore, be calculated as follows:

number of moles = molarity × volume

Volume of KOH = 27.5mL = 27.5/1000

= 0.0275L

n = 0.0275 × 0.250

n = 0.006875 mol

n = 6.88 × 10^-3mol

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A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo

lara [203]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

$\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M$

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

2 years ago

How does a negative ion differ from a positive?

Ilya [14]

A negative ion or anion is an electrically charged atom or group of atoms that are formed by gaining one or more electrons. Whereas a positive ion or cation is an electrically charged atom or group of atoms that are formed by losing one or more electrons.
The losing and gaining all occurs on the outermost shell.

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3 years ago

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evablogger [386]

Answer:

See explanation and image attached

Explanation:

Yttrium has many isotopes, the lowest mass number of Yttrium is 89Y.

Recall that electron capture converts an electron into a proton and then into a neutron with a consequent emission of a neutrino (v).

In electron capture, the mass number of the daughter nucleus remains the same as that of the parent nucleus while the atomic number of the daughter nucleus is less than that of the parent by one unit.

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3 years ago

Identify the main purpose of the Periodic Table. *

saw5 [17]

The periodic table is one of the most important tools in the history of chemistry. It describes the atomic properties of every known chemical element in a concise format, including the atomic number, atomic mass and relationships between the elements. Elements with similar chemical properties are arranged in columns in the periodic table.

The table thus is a quick reference as to what elements may behave the same chemically or which may have similar weights or atomic structures.

Hope this answer helps you

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3 years ago