CHEMISTRY COLLEGE

1. The common name for the compound CH3-CH2-O-CH3 is ​

Answers

Answer 1
Answer:

Answer:

propane

Explanation:

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Answer 2
Answer:

Answer:

Methoxyethane also known as ethyl methyl ether

Related Questions

1‑propanol ( nn ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure and the mole fraction ( yy ) of the vapor phase of each component in equilibrium with each of the given solutions at 25 °C. P∘prop=20.9 TorrPprop°=20.9 Torr and P∘iso=45.2 TorrPiso°=45.2 Torr at 25 °C. A solution with a mole fraction of xprop=0.243xprop=0.243 .
The tides of the oceans and large lakes that occur on Earth are caused by_______The gravitational pull from the moon and sun The earth’s rotation
This is how osmium appears in the periodic table.A purple box has O s at the center and 76 above. Below it says osmium and below that 190.23.Rounded to the nearest whole number, how many neutrons, on average, are in an atom of osmium?76114190266
What is the bond character of this molecule? A.) strongly covalent B.) Positively chargedC.) Strongly ionicD.) Negatively charged
Which response gives the products of hydrolysisofNH4Cl?A. NH4+ + HClB. NH3 + OH- + HClC. NH3 + H+D. NH4OH + HClE. No hydrolysis occurs.

Liquid ethyl mercaptan, C2H6S, has a density of 0.84 g/mL. Assuming that the combustion of this compound produces only CO2 , H2O, and SO2 , what masses of each of these three products would be produced in the combustion of 3.15 mL of ethyl mercaptan

Answers

Answer:

Mass CO2 = 3.75 grams

Mass H2O = 2.30 grams

Mass SO2 = 2.73 grams

Explanation:

Step 1: Data given

Density of Liquid ethyl mercaptan, C2H6S = 0.84 g/mL

Volume of ethyl mercaptan = 3.15 mL

Step 2: The reaction

2C2H6S + 9O2 → 4CO2 + 6H2O + 2SO2

Step 3: Calculate mass of ethyl mercaptan

Mass = Volume * density

Mass ethyl mercaptan = 3.15 mL * 0.84 g/mL

Mass ethyl mercaptan = 2.646 grams

Step 4: Calculate moles ethyl mercaptan

Moles = mass / molar mass

Moles ethyl mercaptan = 2.646 grams / 62.13 g/mol

Moles ethyl mercaptan = 0.04259 moles

Step 5: Calculate moles of other products

For 2 moles ethyl mercaptan we need 9 moles O2 to produce 4 moles CO2, 6 moles H2O and 2 moles SO2

For 0.04259 moles we need 0.1917 moles O2 to produce:

2*0.04259 = 0.08518 moles CO2

3*0.04259 = 0.1278 moles H2O

1*0.04259 = 0.04259 moles SO2

Step 6: Calculate mass produced

Mass = moles * molar mass

Mass CO2 = 0.08518 moles * 44.01 g/mol

Mass CO2 = 3.75 grams

Mass H2O = 0.1278 moles * 18.02 g/mol

Mass H2O = 2.30 grams

Mass SO2 = 0.04259 moles * 64.07 g/mol

Mass SO2 = 2.73 grams

What is anology?
What is microscope?​

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Calculate the mass percent of oxygen in KMnO4.

Answers

Answer: 40.496%

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Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answers

Answer:

\boxed{\text{23.4 mmHg}}

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

K_{\text{p}} = p_{\text{H2O}}

\text{The relationship between $\Delta G^(\circ)$ and $K_{\text{ p}}$ is}\n\Delta G^(\circ) = -RT \ln K_{\text{p}}

Data:  

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

\begin{array}{rcl}8600 & = & -8.314 * 298.15 \ln K \n8600 & = & -2478.8 \ln K\n-3.47 & = & \ln K\nK&=&e^(-3.47)\n& = & 0.0311\end{array}

Standard pressure is 1 bar.

p_{\text{H2O}} = \text{0.0311 bar} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\n\n\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}

A sample of solid calcium hdroxide, Ca(OH)2 is allowed to stand in water until a saturated solution is formed. A titration of 75.00mL of this solution with 5.00 x 10-2 M HCl 36.6 mL of the acid to reach the end pointCa(OH)2 + 2HCl ? CaCl + 2H2O
What is the molarity?

Answers

Answer: The concentration of Ca(OH)_2 is 0.0122 M.

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\nM_1=5.00* 10^(-2)M=0.05M\nV_1=36.6mL\nn_2=2\nM_2=?M\nV_2=75mL

Putting values in above equation, we get:

1* 0.05* 36.6=2* M_2* 75\n\nM_2=0.0122M

Hence, the concentration of Ca(OH)_2 is 0.0122 M.

Which of the following characterizes a reaction at equilibrium?A. The forward reaction stops, and the backward reaction starts.
B. The forward and backward reactions are equal.
C. The forward and backward reactions stop happening.
D. The reactants are changed into products.
SUBMIT

Answers

At chemical equilibrium,the forward and backward reactions are equal and opposite.

What is chemical equilibrium?

Chemical equilibrium is defined as the condition which arises during the course of a reversible chemical reaction with no net change in amount of reactants and products.A reversible chemical reaction is the one wherein the products as soon as they are formed react together to produce back the reactants.

At equilibrium, the two opposing reactions which take place take place at equal rates and there is no net change in amount of the substances which are involved in the chemical reaction.At equilibrium, the reaction is considered to be complete . Conditions which are required for equilibrium are given by quantitative formulation.

Factors which affect chemical equilibrium are change in concentration , change in pressure and temperature and presence of catalyst.

Learn more about chemical equilibrium,here:

brainly.com/question/3920294

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B. The forwards and backwards motions are equal
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