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Vesna [10]
3 years ago
7

A substance is present in the gaseous state at room temperature. Which of the following best explains the probable position of t

he substance in the periodic table?
It is between groups 1 to 12 because it is a metal.
It is between groups 13 to 18 because it is a metal.
It is between groups 1 to 12 because it is a non-metal.
It is between groups 13 to 18 because it is a non-metal.

Chemistry
2 answers:
goldenfox [79]3 years ago
8 0

<u>Answer:</u> The correct answer is it is between groups 13 to 18 because it is a non-metal.

<u>Explanation:</u>

The elements in the periodic table is classified as metals and non-metals.

Metals are the elements which loose electrons. They are present from group 1 to group 12 of the periodic table. They are generally solid at room temperatures. For Example: Sodium, Calcium etc...

Non-metals are the elements which gain electrons. They are present from group 13 to group 18 of the periodic table. They are generally gases at room temperatures. For Example: Fluorine, Oxygen etc..

Hence, the correct answer is it is between groups 13 to 18 because it is a non-metal.

Gnoma [55]3 years ago
3 0

Answer:

D.) It is between groups 13 to 18 because it is a non-metal.

Explanation:

Here we can see the periodic table and what each element is.

It is a non - metal because it is gaseous.

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2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4
agasfer [191]

Answer:

d=4.24x10^{-4}\frac{lb}{in^3}

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

d=\frac{m}{V}

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

m=552.4g-464.7g=87.7g

So that we are now able to calculate the density in g/mL first:

d=\frac{87.7g}{27.8mL}=3.15g/mL

Now, we proceed to the conversion to lb/in³ by using the following setup:

d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}

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6 0
2 years ago
Aqueous hydrobromic acid hbr will react with solid sodium hydroxide naoh to produce aqueous sodium bromide nabr and liquid water
Natalka [10]
 Balanced equation is

HBr + NaOH ----> NaBr + H2O

Using molar masses

80.912 g HBr reacts with  39.997 g of Naoh to give 18.007 g water

so 1 gram of NaOH reacts with 2.023 g of HBR   
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction

Mass  of water produced  =    (5.7 * 18.007 / 39.997  =  2.6 g to 2 sig figs
8 0
3 years ago
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A. is the answer
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2 years ago
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When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb
aliya0001 [1]

The heat released in the combustion of lactic acid is absorbed by the

calorimeter and in the decomposition of the lactic acid.

ΔH°f of lactic acid is approximately <u>-716.2 kJ</u>

Reasons:

Known parameters are;

Mass of the lactic acid = 3.93 grams

Heat  capacity of the bomb calorimeter = 10.80 kJ·K⁻¹

Change in temperature of the calorimeter, ΔT = 5.34 K

ΔHrxn = ΔErxn

ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹

ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹

The chemical equation for the reaction is presented as follows;

  • C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O

The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ

Molar mass of C₃H₆O₃ = 90.07 g/mol

Number of moles of C₃H₆O₃ = \dfrac{3.93 \, g}{90.07 \, g/mol} = 0.043633 moles

Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles

Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ

Moles of H₂O produced = 0.130899 moles

Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ

Therefore, we have;

Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter

Which gives;

Heat absorbed by lactic acid  = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ

The heat absorbed by the lactic acid ≈ -31.249 kJ

  • \Delta H^{\circ}f \ of \ C_3H_6O_3 = \dfrac{-31.249}{0.043633} \approx  -716.2

ΔH°f of C₃H₆O₃ ≈ -716.2 kJ

Heat of formation of lactic acid ≈ <u>-716.2 kJ</u>.

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