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4 years ago

5

In an experiment you are asked to react hydrochloric acid with sodium hydroxide. when measuring the volume of the reactants, whi

ch instrument would give the greatest precision?

2 answers:

AveGali [126]4 years ago

5 0

Answer;

- A 100 mL graduated cylinder since it has marks showing the volume of the substance measured.

Explanation;

-A graduated cylinder or a measuring cylinder is a common piece of laboratory equipment used to measure precise volumes of a liquid.

-Graduated cylinders are specifically designed to measure out liquid volumes. Their tall narrow design makes for a more precise reading of the liquid level as compared to other pieces used to measure volume.

melomori [17]4 years ago

3 0

When measuring the volume of the reactants, the instrument that <span>would give the greatest precision would be a graduated cylinder since it has marks showing the volume of the substance measured. Hope this answers the question. Have a nice day.</span>

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Identify the decay mode of hydrogen -3.

Phantasy [73]

Answer:

The radioactive decay mode that does not result in a different nuclide is gamma decay. In gamma decay, the only product is a gamma ray photon.

Explanation:

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2 years ago

Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr

Ganezh [65]

Answer:

a. $4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas $N_2$ , oxygen gas $O_2$ , water vapor $H_2O$ and carbon dioxide $CO_2$ . Let's write the decomposition of nitroglycerin into these 4 components:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)$

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by $\frac{3}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by $\frac{5}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

$\frac{5}{2} + 6 = 8.5$

This leaves $9 - 8.5 = 0.5 = \frac{1}{2}$ of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

To make it look neater without fractional coefficients, multiply both sides by 4:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

$V_{CO_2} = 41.0 L$

$T = -14.0^oC + 273.15 K = 259.15 K$

$p = 1 atm$

Firstly, we may find moles of carbon dioxide produced using the ideal gas law $pV = nRT$ .

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

$n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol$

According to the stoichiometry of the balanced chemical equation:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

$\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol$

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

$m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g$

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4 years ago

Plzz some one tell me tha objection 5 6 points to kekule formula

kakasveta [241]

A structural formula for an organic compound that depicts each valence bond as a short line especially : the hexagonal ring formula for benzene

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3 years ago

Which of the following molecular compounds does NOT need the usage of prefix for the first element

Mrrafil [7]

Answer:

SO2

Explanation:

that's my answer

Sana mKatolong

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2 years ago

It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize w

Verdich [7]

The question is incomplete and complete question is :

It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.050 mL. Assuming the density of liquid water is 1.00 g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100°C and 1atm ?

Answer:

0.0028 moles of water are in one drop of liquid.

Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.

Explanation:

Volume of of drop = v = 0.050 mL

Mass of drop = m

Density of water = d = 1.00 g/mL

$m=d\times v=0.050 mL\times 1.00 g/mL=0.050 g$

Moles of water in drops:

$\frac{0.050 g}{18 g/mol}=0.0028 mol$

0.0028 moles of water are in one drop of liquid.

Pressure at which 0.0028 moles are vaporized = P = 1 atm

Temperature at which 0.0028 moles are vaporized = T = 100°C = 100+273 K = 373 K atm

Volume of moles of water = V

Moles of water = n = 0.0028 mol

$PV=nRT$ ( ideal gas equation )

$V=\frac{0.0028 mol\times 0.0821 atm L/mol K\times 373 K}{1 atm}$

V = 0.086 L = 86 mL ( 1L = 1000 mL)

Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.

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3 years ago