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2 years ago

6

A teacher wants to figure out The amount of a solution that is needed for a titration experiment in a class lab! the class has 1

5 students in each experiment needs 50 mL of solution. if each student will perform the experiment two times how many milliliters of a solution will the teacher need for the whole class

2 answers:

AnnZ [28]2 years ago

5 0

The teacher needs 1500 mL for the whole class.

Anastaziya [24]2 years ago

3 0

<u>Answer:</u> Teacher would require 1500mL of solution for the whole class.

<u>Explanation:</u>

It is given that 50mL of a solution is required by 1 student to perform 1 set of titration and each student does the experiment 2 times, so every student would require (50 × 2) = 100mL of the solution.

There are in total 15 students in the class, so total volume of the solution required by the class will be = ( 15 × 100) = 1500mL

Hence, teacher would require 1500mL of solution for the whole class.

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Draw the diazonium cation formed when cytosine reacts with nano2 in the presence of hcl.

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Answer in the Word document below.
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2 years ago

A salt is an ionic compound containing cations other than hydrogen and anions other than the hydroxylion. in writing the formula

Blizzard [7]

In writing the formula for a salt the symbol of the cation is first then the anion is written second .

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2 years ago

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

3 0

2 years ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0

1 year ago

A student performed an analysis of a sample for its calcium content and obtained the following results:

sweet-ann [11.9K]

Answer:

14.9075 g, 28.67%, 0.11%

Explanation:

The mean concentration of calcium = summation x / frequency

= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g

Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164

b) percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%

c) relative standard deviation = standard deviation / mean × 100 = 0.0164 / 14.9075 × 100 = 0.11%

d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.

The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.

5 0

2 years ago