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rosijanka [135]
3 years ago
15

Determine the length of the object shown. 7 cm 7.2 cm 72.5 mm 72.50 mm

Chemistry
2 answers:
ANTONII [103]3 years ago
6 0

Answer:

7.2!!!!

Explanation:

jekas [21]3 years ago
5 0

Answer:

either 7.2 or 7.25

Explanation: its not 7 because it starts before 1 not on 1 (i cant remember the answer but sorry if its wrong

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Matter is never created or destroyed.
kumpel [21]
I believe it is true!
8 0
3 years ago
A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed
Archy [21]

Answer:

The molar mass of the acid is 167.5 g/mol

Explanation:

A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed a single end point after 19.31 mL of 0.1243 M NaOH had been introduced. Calculate the molecular mass of the acid.

Step 1: Data given

Mass of the sample of a purified organic acid = 0.4021 grams

Molarity = 0.1243 M

Volume needed to reach the end point = 19.1 mL = 0.01931 L

Step 2: Calculate the number of moles NaOH

Moles NaOH = molarity NaOH  * volume

Moles NaOH = 0.1243 M * 0.01931 L

Moles NaOH = 0.00240 moles

Step 3: Calculate moles of the acid

We'll need 0.00240 moles of acid to neutralize 0.00240 moles of NaOH ( it's a single end point)

Moles acid = 0.00240 moles

Step 4: Calculate molar mass of the acid

Molar mass = mass / moles

Molar mass = 0.4021 grams / 0.00240 moles

Molar mass = 167.5 g/mol

The molar mass of the acid is 167.5 g/mol

6 0
3 years ago
2.(04.01 LC)
Mila [183]

Answer:

2KClO3 —> 2KCl + 3O2

The coefficients are 2, 2, 3

Explanation:

From the question given above, we obtained the following equation:

KClO3 —> 2KCl + 3O2

The above equation can be balance as follow:

There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:

2KClO3 —> 2KCl + 3O2

Now, the equation is balanced.

Thus, the coefficients are 2, 2, 3

8 0
3 years ago
Convert 35.0 m/s to cm/min
swat32
35 m/s = 210000 cm/min
35*600=210000
3 0
3 years ago
What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?
KonstantinChe [14]

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird=  11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that  

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

5 0
3 years ago
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