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3 years ago

Geiger counters and scintillation counters differ in Blank .

Chemistry

2 answers:

AfilCa [17]3 years ago

4 0

Answer:

Geiger counters can detect all kinds of radiation, while scintillation counters can only detect ionizin radiation

butalik [34]3 years ago

3 0

The Geiger Counter. Geiger counters are used to detect radioactive emissions, most commonly beta particles and gamma rays. The counter consists of a tube filled with an inert gas that becomes conductive of electricity when it is impacted by a high-energy particle.

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Write the complete balanced equation for the decomposition of AI(CIO3)3.

OlgaM077 [116]

Answer:

2AL(ClO3)3 → 2ALCl3 + 9O2

Explanation:

3 0

3 years ago

Surrounding ions with water molecules is known as

Stells [14]

Answer:

Hydration

Explanation:

In water, ions are surrounded by a sphere of water molecules called a hydration shell. The process of forming this shell is called hydration.

3 0

3 years ago

match the number of negative charges to the number of positive charges to make each combination balanced (see image for answer)

BartSMP [9]

Answer : The correct match is:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

Explanation :

As we now that there are three subatomic particles which are protons, electrons and neutrons.

The protons and neutrons are located inside the nucleus and electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and neutrons are neutral.

As we know that all the things are made up of charges and opposite charges attract to each other.

In a neutral atom, the positive charges and negative charges are balanced in an object. That means, in neutral atom the number of positive charges are equal to the negative charges.

So we can say that:

1 positive charge = 1 negative charge

2 positive charges = 2 negative charges

3 positive charges = 3 negative charges

5 0

4 years ago

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .