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Ugo [173]
3 years ago
9

What is the balenced formula for C3H8+O2

Chemistry
1 answer:
balandron [24]3 years ago
4 0

C3H8  +   5O2  ----->   3CO2  +   4H2O

This is propane  burning in air or pure oxygen.

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What is/are the principal product(s) formed when excess methylmagnesium iodide reacts with p-hydroxyacetophenone?
Montano1993 [528]

The end product will depend upon

a) the amount of the reagent taken

b) the final treatment of the reaction

If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen

Figure 1

However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product

Figure 2

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3 years ago
Chemical formula for 2 butyene is <br>​
yarga [219]

Answer:

\text{C}_4\text{H}_6

Explanation:

3 0
2 years ago
I am a gas. I am in the noble gas family and row 1. I have only 2 valence electrons I glow red-orange when in an electric field.
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7 0
2 years ago
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When 8.0 grams of sodium hydroxide is dissolved in sufficient water to make 400. mL of solution, what is the concentration of th
nata0808 [166]

Answer:

The concentration of this sodiumhydroxide solutions is 0.50 M

Explanation:

Step 1: Data given

Mass of sodium hydroxide (NaOh) = 8.0 grams

Molar mass of sodium hydroxide = 40.0 g/mol

Volume water = 400 mL  = 0.400 L

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 8.0 grams / 40.0 g/mol

Moles NaOh = 0.20 moles

Step 3: Calculate concentration of the solution

Concentration solution = moles NaOH / volume water

Concentration solution = 0.20 moles / 0.400 L

Concentration solution = 0.50 M

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6 0
3 years ago
Glacier National Park in Montana is approximately 4,100 ft above sea level with an atmospheric pressure of 681 torr. At what tem
lapo4ka [179]

Answer : The temperature of liquid is, 369.9 K

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of liquid at 373 K = 681 torr

P_2 = vapor pressure of liquid at normal boiling point = 760 torr

T_1 = temperature of liquid = ?

T_2 = normal boiling point of liquid = 373 K

\Delta H_{vap} = heat of vaporization = 40.7 kJ/mole = 40700 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760torr}{681torr})=\frac{40700J/mole}{8.314J/K.mole}\times (\frac{1}{T_1}-\frac{1}{373K})

T_1=369.907K\approx 369.9K

Hence, the temperature of liquid is, 369.9 K

6 0
2 years ago
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