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2 years ago

The Earth's orbit is a(n) ___shape. .

Chemistry

1 answer:

Stels [109]2 years ago

8 0

Answer:

elliptical/ellipse

Explanation:

» Concepts

Astronomer Johannes Kepler discovered and founded his three laws of planetary motion. One of these laws states that the orbit of all planets, including Earth, is in an elliptical shape rather than a circle.

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What’a the structure of 4-ethyl-2,2-dimethylheptane?

Slav-nsk [51]

PubChem CID: 142982

Chemical Names: 4-Ethyl-2,2-dimethylhexane; Hexane, 4-ethyl-2,2-dimethyl-; 52896-99-8; 2,2-dimethyl-4-ethylhexane; AC1L3M80; Hexane,4-ethyl-2,2-dimethyl- More...

Molecular Formula: C10H22

Molecular Weight: 142.286 g/mol

InChI Key: QHLDBFLIDFTHQI-UHFFFAOYSA-N

7 0

2 years ago

2. The empirical formula of a molecule is CH2O. In an experiment, the molar mass of the molecule was determined to be 360.3 g/mo

Vanyuwa [196]

Answer:

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

x = 360.3/30

x = 12

Finally

C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation

2Mg + O2 ⇒ 2MgO

48.62 g of Mg ----------------- 80.62 g of MgO

35g ------------------ x

x = 58 g of MgO (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

= 90 x 58 / 100

= 52. 23 g

5 0

3 years ago

Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.

Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

To find the number of moles:

number of moles = $\frac{number of particles}{Avogadro's number}$

number of moles = $\frac{6.29 x 10^{23} }{6.02 x 10^{23} }$

number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

#learnwithBrainly

7 0

2 years ago

How many particles are in 697.008 grams of Potassium Sulfate?

g100num [7]

Answer:

174.2592

Explanation:

Trust me.

8 0

2 years ago

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0

3 years ago