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3 years ago

Is it a plant or animal cell?

Chemistry

1 answer:

Charra [1.4K]3 years ago

8 0

i think there was ment to be a picture here

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What is the percent by volume of a solution formed by mixing 15 mL of hydrochloric acid into water, forming 75 mL of solution?

timama [110]

Answer:

= 17% (v/v)

Explanation:

% by Volume = (Volume HCl / Volume of Solution)100%

= 15ml/(75ml + 15ml) x 100% = 16.66666667% (calculator answer)

= 17% (v/v) 2 sig figs. based on volume values given.

4 0

3 years ago

Calculate the hydroxyl ion concentration<br>in 0.02M solution of H2SO4<br>

sergij07 [2.7K]

In a given solution, the concentration of hydrogen ions and hydroxide ions are related to each other by the following expression:

pH + pOH = 14

where, pH = - log[H+] and pOH = - log [OH-]

In this case, the concentration of sulfuric acid is given as 2.1 x 10^-4 M. Since each molecule of sulfuric acid (H2SO4) contains 2 atoms of hydrogen, the concentration of hydrogen ions in this solution is twice that of sulfuric acid. That is,

Concentration of H+ ions = 2 x 2.1 x 10^-4

= 4.2 x 10^-4 M

This means, pH = -log (4.2 x 10^-4) = 3.38

Since pH + pOH = 14

pOH = 14 - 3.38 = 10.62

This means, 10.62 = -log [OH-]

Solving the equation, we get [OH-] = 2.4 x 10^-11 M.

Hope this helps.

7 0

4 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago

How many grams are in .705 moles of Cr?

Kruka [31]

Answer: 51.9961 grams.

Explanation:

3 0

3 years ago

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There are about 1 760 000 000 000 000 000 000 molecules of sucrose in 1 g of table sugar. Expressed in scientific notation, what

Dmitriy789 [7]

The correct answer from the given choices is A. One gram of table sugar is equal to 1.76 x 10^21 molecules of sucrose. Scientific notation is a method that scientists use to easily handle numbers that are very large or very small.

8 0

4 years ago

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