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Inessa05 [86]
3 years ago
15

Balance CH3CHO + O2 → CO2 + H2O

Chemistry
1 answer:
babunello [35]3 years ago
7 0
2 CH3CHO + 5 O2 = 4 CO2 + 4 H2O

reaction type = combustion

hope this helped :)

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Which two atoms could be the "before'' and ''after'' of an alpha ejection
spin [16.1K]
The answer would be uranium and thorium. When an alpha ejects a particle, it will create a new atom. So, when uranium ejects an alpha particle, it will produce thorium.  They call this process as the alpha decay. Alpha decay often happens on atoms that are abundant nuclei such as uranium, radium, and thorium.
4 0
3 years ago
When does boiling occur?
butalik [34]

Water boils at 100 Degrees Celsius or 212 degrees Fahrenheit

8 0
4 years ago
What common item can detect the presence of radiation? A. Bleach B. A roll of photographic film C. LED watch D. Baking soda
Alenkinab [10]

Answer: B- a roll of photographic film

Explanation:

Founders education answer

6 0
3 years ago
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If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
How many aluminum atoms are in 3.78g of aluminum?
KiRa [710]

Answer:

The molar mass of aluminum is 26.9 grams per mole. We start by converting the grams of aluminum to moles of aluminum.

m

o

l

 

A

l

 

=

 

3.78

 

g

 

A

l

×

1

 

m

o

l

 

A

l

26.9

 

g

 

=

 

0.1405

 

m

o

l

 

A

l

We use Avogadro's number to determine the number of aluminum atoms present in 0.1405 mol Al.

A

t

o

m

s

 

A

l

 

=

 

0.1405

 

m

o

l

 

A

l

×

6.022

×

10

23

1

 

m

o

l

 

A

l

 

=

 

8.46

×

10

22

 

a

t

o

m

s

 

A

l

8 0
3 years ago
Read 2 more answers
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