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3 years ago

There is a container on the lab bench containing some unknown chemical. You should:

Chemistry

2 answers:

Elis [28]3 years ago

5 0

Answer:

Explanation:

Your supervisor is in charge of you so therefore you must always tell your supervisor first.

lesantik [10]3 years ago

3 0

Answer:

Explanation:

Because you might get hurt, could be dangerous.

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Complete the following paragraph.

gladu [14]

Explanation:

the stationary phase in TLC is a silica gel coated metal plate or paper.

The individual component of the mixure will travel a greater distance up the plate, resulting in a Rf value for the component.

the p-Xylene spot will travel a greater distance as it is the lighter component

Bromobenzene will travel least as it is the heavier component.

thin layer chromatography (TLC) is a type of chromatographic seperation technique that is based on the molecular size of components. the Rf value is the distance covered by the component relative to that travelled by the solvent which is the mobile phase

3 0

4 years ago

The freezing point (tf) for t-butanol is 25.50°c and kf is 9.1°c/m. Usually t-butanol absorbs water on exposure to the air. If t

padilas [110]

The mathematical expression is given as:

$\Delta T = k_{f}m$ (1)

where, $\Delta T$ = depression in freezing point

$k_{f}$ = molal freezing point.

Now, first calculate the $\Delta T = 25.50^{o}C -24.59^{o}C$

$\Delta T= 0.91^{o}C$

Substitute the values in equation (1), we get

$0.91^{o}C = 9.1^{o}C/m \times m$

$m = \frac{0.91^{o}C}{9.1^{o}C/m}$

= $0.1 molal$ or $\frac{0.1 moles of water}{kg of butanol}$

Now,

In 0.1 moles of water = 1 kg of butanol

So, 11.9 g of butanol = $11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}$

Convert gram into kilogram, (1 kg =1000 g)

= $11.9 g butanol\times \frac{0.1 mol of water}{kg butanol}\times \frac{1 kg}{1000 g}$

= $0.00119 mole$

Mass of water present in sample = $0.00119 mole\times 18 g/mol$

= $0.02142 g$

Hence, grams of water present in the sample = $0.02142 g$

8 0

3 years ago

How much energy is produced in the creation of 5 grams of "O by the process: (10 pts.) 14N + α. 'H + 170 ("N-14.00307 gmole, α-4

LenKa [72]

Answer : The energy produced is $3.410\times 10^{10}J$

Explanation :

First we have to calculate the moles of $^{17}O$ .

$\text{Moles of }^{17}O=\frac{\text{Mass of }^{17}O}{\text{Molar mass of }^{17}O}=\frac{5g}{16.99913g/mole}=0.294133moles$

Now we have to calculate the mass defect.

The balanced reaction is,

$^{14}N+\alpha \rightarrow ^1H+^{17}O$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=[(n_{^1H}\times M_{^1H})+(n_{^{17}O}\times M_{^{17}O})]-[(n_{^{14}N}\times M_{^{14}N})+(n_{\alpha}\times M_{\alpha})]$

where,

n = number of moles = 0.294133 moles

M = molar mass

Now put all the given values in the above, we get:

$\Delta m=[(n_{^1H}\times M_{^1H})+(n_{^{17}O}\times M_{^{17}O})]-[(n_{^{14}N}\times M_{^{14}N})+(n_{\alpha}\times M_{\alpha})]$

$\Delta m=[(0.294133mole\times 1.00783g/mole)+(0.294133mole\times 16.99913g/mole)]-[(0.294133mole\times 14.00307g/mole)+(0.294133mole\times 4.0026g/mole)]$

$\Delta m=0.00037943157g=3.7943157\times 10^{-7}kg$

Now we have to calculate the energy produced.

$Energy=\Delta m\times (c)^2$

$Energy=(3.7943157\times 10^{-7}kg)\times (299792458m/s)^2$

$Energy=3.410\times 10^{10}J$

Therefore, the energy produced is $3.410\times 10^{10}J$

3 0

4 years ago

reactant → product If there are 4 grams of reactant, how many grams of product are produced by the chemical reaction? A) 0

m_a_m_a [10]

Answer: Option c is correct.

Explanation:

According to the law of conservation of mass or matter for a chemical reaction states that matter can neither be created nor be destroyed. It can simply be transformed from one form to another.

Therefore, if there are 4 grams of reactant then there will be production of 4 grams of product.

Thus, it is concluded that option c that is 4 grams is the correct answer.

6 0

3 years ago

Read 2 more answers

18. Which two particle diagrams each represent a sample of one substance? Key O = atom of oxygen = atom of hydrogen 0 = atom of

Mars2501 [29]

Answer:

1 and 11

Explanation:

3 0

3 years ago