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asambeis [7]
3 years ago
5

Who made the periodic table?

Chemistry
2 answers:
swat323 years ago
7 0
Nobody specifically made the periodic table but it was first thought of by Dmitri Mendeleev in 1869.
DENIUS [597]3 years ago
6 0
Dmitri Mendeleev invented the periodic table.
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a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find t
dexar [7]

Answer:

empirical formula = C3H7

molecular formula = C6H14

3 0
3 years ago
What happens to an atom if it loses an electron?
jolli1 [7]

Answer:

When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons. In the opposite case when an atom gains electron(s) it becomes negatively charged (more electrons than protons).

Explanation:

7 0
3 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
Acetylene is hyrodgenated to form ethane. The feed to the reactor contains 1.60 mol H2/mol C2H2. The reaction proceeds to comple
Black_prince [1.1K]

Answer:

Explanation:

C₂H₂     +   2H₂   =     C₂H₆

1 mole     2 mole        1 mole

Feed of reactant is 1.6 mole H₂ / mole C₂H₂

or 1.6 mole of H₂ for 1 mole of C₂H₂

required ratio as per chemical reaction written above

2 mole of  H₂ for 1 mole of C₂H₂

So H₂ is in short supply . Hence it is limiting reagent .

1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆

a )Calculate the stoichiometric reactant ratio =  mole H₂ reacted/mole C₂H₂ reacted

= 1.6 / .8 = 2 .

b )

yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .

4 0
3 years ago
When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac
slega [8]
<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0
1 year ago
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