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Lilit [14]
3 years ago
9

What is the name of this compound? OCH3CCH2 CH3

Chemistry
2 answers:
ELEN [110]3 years ago
7 0
Simple really, when you...
Maurinko [17]3 years ago
6 0
From question to your answer is
butanal
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What is difference between ∆H and ∆H°
Elena-2011 [213]

If it is shown as ∆H , then it means that a specific chemical reaction is undergoing heat in Kelvin(K). If it is shown as ∆H° , then it means that a specific chemical reaction is undergoing heat in Celsius(C⁰).

Hope this is the answer you are looking for mate!

If it is correct then please mark my answer as the brailiest! :)

5 0
3 years ago
This question has two parts btw
Oxana [17]

Answer:

the cold one is going to have a faster reaction time

Explanation:

5 0
3 years ago
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Mercury (Hg) poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes l
alisha [4.7K]

Answer:

6 x 10⁵ kg Hg

Explanation:

The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.

The volume of the lake is calculated in cubic feet:

V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³

Cubic feet are then converted to mL (1cm³=1mL)

(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL

The mass of mercury is then found:

m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg

3 0
3 years ago
The molar heat of vaporization of hydrogen fluoride (HF) is 31.18 kJ/mol. How much heat is released when 100.0 g HF condenses at
mash [69]

the answer is -155.8


8 0
3 years ago
Read 2 more answers
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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