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lawyer [7]
3 years ago
13

Which of the following is an empirical formula? OP205 O C2H602 ON3F6 OC8H18

Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer:

None are empirical formulas

Explanation:

All are actual compounds. An example of an empirical formula could be CH2O, the empirical formula for carbohydrates like glucose (C6H12O6).

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Type the correct answer in the box.
Murljashka [212]

Answer:

<em>a= In scientific notation</em>

6.96×10⁵ Km

<em>b =In  expanded notation</em>

0.00019 mm

Explanation:

Given data:

Radius of sun = 696000 Km

size of bacterial cell = 1.9 ×10⁻⁴ mm

Radius of sun in scientific notation = ?

Size of bacterial cell in expanded notation = ?

Solution:

Radius of sun:

696000 Km

<em>In scientific notation</em>

6.96×10⁵ Km

Size of bacterial cell:

1.9 ×10⁻⁴ mm

<em>In  expanded notation</em>

1.9/ 10000 = 0.00019 mm

8 0
3 years ago
The charge on a magnesium ion is +2. When it formed an ionic compound with another element, it lost
strojnjashka [21]
I believe it’s loses two electrons.
4 0
3 years ago
Convert 7.50 grams of glucose C6H12O6 to moles
Lera25 [3.4K]

Answer:

The number of mole is 0.04167mole

Explanation:

To convert gram to mole, we need to calculate the molecular weight of the compound

C6H12O6

C - 12

H - 1

O - 16

Molecular weight = 6 * 12 + 1 *12 + 6 * 16

= 72 + 12 + 96

= 180g/mol

To covert gram to mole

Therefore,

= 7.50g/ 180g/mol

= 0.04167 mole of glucose

3 0
3 years ago
Why do you think scientist can use the basic properties of matter to help identify an unknown substance?
deff fn [24]

They can use the properties to test and come to some kind of conslusion about the object because in some way it's gotta correlate back to one of the different properties of matter.

4 0
3 years ago
A sample contains 25% parent isotope and 75% daughter isotopes. If the half-life of the parent isotope is 72 years, how old is t
alisha [4.7K]

The radioactive decay obeys first order kinetics

the rate law expression for radioactive decay is

ln\frac{[A_{0}]}{[A_{t}]}=kt

Where

A0 = initial concentration

At = concentration after time "t"

t = time

k = rate constant

For first order reaction the relation between rate constant and half life is:

k=\frac{0.693}{t_{\frac{1}{2} } }

Let us calculate k

k = 0.693 / 72 = 0.009625 years⁻¹

Given

At = 0.25 A0

ln(\frac{A0}{0.25A0})=0.009625  X time

time = 144 years

So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**

Simply two half lives

5 0
3 years ago
Read 2 more answers
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