Convert 175 lbs to kg (1 kg = 2.2 lbs)

Which radioactive waste can be stored for decay and then safely released into the environment?

Natalka [10]

The radioactive waste can be stored for decay and then safely released into the environment is N-16.

<h3>What are the types of decay?</h3>

Decays can be of three types: alpha, beta and gamma. Each of them corresponds to a different radioactive particle, which changes the nucleus of the emitting atom according to its characteristics.

Nitrogen 16 ( 16 N ) is the unstable isotope of nitrogen whose nucleus consists of 7 protons and 9 neutrons. Its period is 7.13 s.

See more about nitrogen at brainly.com/question/16711904

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4 0

2 years ago

[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con

telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = $284 \ grams$

solid soil volume = $205 \ cc$

saturated mass soil = $361 \ g$

The weight of the soil after drainage is = $295 \ g$

Water weight for soil saturation = $(361-284) = 77 \ g$

Water volume required for soil saturation = $\frac{77}{1} = 77 \ cc$

Sample volume of water: $= \frac{\text{water density}}{\text{water density input}}$

$= 361- 295 \\\\ = 66 \ cc$

Soil water retained volume = (draining field weight - dry soil weight)

$= 295 - 284 \\\\ = 11 \ cc.$

$\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}$

$= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%$

(Its saturated water volume is equal to the volume of voids)

$\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}$

$= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23$

$\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}$

$= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04$

6 0

3 years ago

65.39 Atomic # = Atomic Mass = # of Protons = # of Neutrons = # of Electrons =

ratelena [41]

Answer:

The answer to your question is below

Explanation:

Atomic mass = 65.39 g

Searching this number in the periodic table we find that the element is Zinc.

Then:

# of Protons = 30

# of neutrons = atomic mass - # of protons

= 65.39 - 30

= 35.39

# of electrons = # of protons = 30

8 0

4 years ago

Approximately how many moles of chlorine make up 1.79 × 1023 atoms of chlorine?

e-lub [12.9K]

1.2*10^24# atoms of chlorine

Explanation:
Chlorine gas (#Cl_2#) has two atoms of elemental chlorine in a molecule, so:
#1# mol of #Cl_2# have #6*10^23# molecules of #Cl_2#
#1# molecule of #Cl_2# have #2# atoms per molucule
Then #2*6*10^23 = 1.2*10^24# atoms of chlorine in a mol of chlorine gas

6 0

3 years ago

A sample of sodium sulfite has a mass of 2.80 g.

Gnom [1K]

Answer:

For a: The number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b: The number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c: The mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

Explanation:

The chemical formula of sodium sulfite is $Na_2SO_3$ . It is formed by the combination of 2 sodium $(Na^+)$ ions and 1 sulfite $(SO_3^{2-})$ ions

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

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

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol$

For a:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22}$ number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b:

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22}$ number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c:

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of formula units are present in 1 mole of a compound

Or, $6.022\times 10^{23}$ number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = $\frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g$

Hence, the mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

7 0

3 years ago