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3 years ago

Convert 0.30 m to mm.

Chemistry

1 answer:

scoundrel [369]3 years ago

7 0

Answer:

0.3 Meters = 300 Millimeters

Explanation:

Multiply the length value by 1000

Hopefully, this helps! :D

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Construct a three-step synthesis of 1,2-epoxycyclopentane from cyclopentanol by dragging the appropriate formulas into the bins.

zubka84 [21]

Answer:

(1) Bromination, (2) E2 elimination and (3) epoxidation

Explanation:

In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with $PBr_{3}$
In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
Full reaction scheme has been shown below

3 0

3 years ago

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

Answer: The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

3 years ago

Mixtures of hydrogen and oxygen in varying concentrations can be sparked to produce water. Define mixture and compound, and then

LenKa [72]

Explanation:

Mixture is the physical Combination Of two or Substance

Example

a mixture of sugar and water.

Compound is the chemical combination of two or more metals.

Example.

a mixture of hydrogen and water.

A mixture of hydrogen and oxygen forms water or H2O

whereas The dihydrogen monoxide parody involves calling water by an unfamiliar chemical name, most often "dihydrogen monoxide" (DHMO), and listing some of water's properties in a particularly alarming manner, such as accelerating corrosion (rust) and causing suffocation (drowning). The parody often calls for dihydrogen monoxide to be banned, regulated, or labeled as dangerous. It plays into chemophobia and demonstrates how a lack of scientific literacy and an exaggerated analysis can lead to misplaced fears. The parody has been used with other chemical names such as hydrogen hydroxide, dihydrogen oxide, hydroxic acid, hydric acid and oxidane.

4 0

3 years ago

In an atomic model that includes a nucleus, positive charge is

olganol [36]

Because protons are positively charged and neutrons have no charge then it is safe to say that such an atomic model would have the positive charge concentrated in the center of an atom (option d).

7 0

3 years ago

A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

4 years ago