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3 years ago

Hydrogen is an elemental gas why

Chemistry

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Answer:

hydrogen is a colorless, odorless, nonmetallic, tasteless, highly flammable diatomic gas

Explanation:

Hydrogen is part of the Noble gas family

meriva3 years ago

5 0

Answer: Hydrogen is an elemental gas because it is composed of a single proton and a single electron.

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2.714 mole of helium has a mass of ______ grams. (Enter just the number for this one.)

kifflom [539]

Hello!

To find the mass of helium, we need to multiply the total moles by the mass of helium. We are given 2.714 moles of helium, and the mass of helium is about 4.00 grams. Now, we multiply the two values together to get the grams.

2.714 moles x 4.00 grams = 10.856 grams

According to the number of significant figures, 2.714 moles of helium has a mass of 10.9 grams (exact value: 10.856 grams).

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3 years ago

Explain why the equator has a warmer climate than the area near either the North or South pole. Be sure to thoroughly explain yo

KatRina [158]

Il fait plus chaud à l'équateur et plus froid aux pôles car : 1°) les rayons du Soleil sont plus concentrés au niveau de l'équateur et plus diffus au niveau des pôles ; 2°) l'épaisseur d'air composant l'atmosphère, traversée par les rayons du Soleil est plus importante aux pôles qu'à l'équateur.

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2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Students conduct an experiment:

sweet-ann [11.9K]

Answer:

The balloon becomes inflated

Explanation:

The equation of the reaction between baking soda (sodium bicarbonate) and vinegar(ethanoic acid) is shown below;

NaHCO3 + HC2CH3O2 ------> NaC2H3O2 + H2O + CO2

The gas (CO2) evolved in the process leads to the inflation of the balloon dropped on the bottle in which the reaction is taking pace.

This observation provides evidence that a gas was really evolved in the reaction.

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3 years ago

Grade 8 Science

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Materials and their properties: compounds like sodium chloride - an interactive educational resource for 11 to 14 year olds. ... Elements are substances (like hydrogen and oxygen) that can't be split into simpler substances. ... For each statement, decide whether it describes a mixture or a compound and check the box.

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3 years ago

Hydrogen is an elemental gas why​

Hydrogen is an elemental gas why