No, because their polarities are too far apart. You can circumvent this by adding a solvent that has a polarity in between these two solvents, e.g. ethanol, methanol, dioxane, dimethylformamide or dimethylsulfoxide.
No the two are immiscible and trichloroethane is denser than water
Yes, H20 and CH3OH are miscible.
Co2 + 3h2 gives ch3oh + h2o
1: Yes, Two polar compounds are miscible. 2. No, CBr4 is non polar, while H2O is polar; immiscible. 3. No, Cl2 is non polar, while H2O is polar: immiscible.
a. KCl in water b. CH3OH in H2O c. CH2Cl2 in benzene
CuO + CH3OH --> HCHO + Cu + H2O
H2O is very slightly soluble in CH2CL2 but H2O is hydrophillic and CH2Cl2 is hydrophobic therefore the two molecules dont react or bond properly resulting in a two phase solution of an aquous and organic layer containg a very small percentage of the H2O.
Ch3oh
Co2 + 3h2 gives ch3oh + h2o
No. Hydrocarbons are not miscible in water.
1: Yes, Two polar compounds are miscible. 2. No, CBr4 is non polar, while H2O is polar; immiscible. 3. No, Cl2 is non polar, while H2O is polar: immiscible.
Methanol (CH3OH) reacts with water (H2O) to form methyl alcohol (CH3OH) and a hydroxide ion (OH-). The balanced equation for this reaction is: CH3OH + H2O -> CH3OH + OH-
a. KCl in water b. CH3OH in H2O c. CH2Cl2 in benzene
CuO + CH3OH --> HCHO + Cu + H2O
As with any combustion reaction you need to include oxygen. The full equation for methanol combustion is: CH3OH + 2 O2 --> CO2 + 2 H2O
When methanol (CH3OH) reacts with sulfuric acid (H2SO4), it forms dimethyl ether (CH3OCH3) and water (H2O) as products. This reaction is known as the dehydration of methanol.
its balanced
To calculate the amount of NH3 needed to react with 21 grams of CH3OH, you first need to balance the chemical equation for the reaction. Then, you convert the mass of CH3OH to moles and use the mole ratio from the balanced equation to determine the moles of NH3 required. Finally, convert the moles of NH3 to grams using the molar mass of NH3.
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH