All categories

1 year ago

Are the properties of a specific element the same as the properties of a molecule made up of that element? Why or why not?

Chemistry

1 answer:

LuckyWell [14K]1 year ago

4 0

Answer:

no molecule and properties or defent

You might be interested in

Look at the picture

Ronch [10]

Answer:

The strong electric force.

Explanation:

Trust me. If its wrong, comment on this, don't delete it. If someone else answers with the wrong question than other students might get the answer wrong. So if its right, say that it is, if its wrong, tell others what the real answer is.

3 0

1 year 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

2 years ago

Can anyone help me with this?

satela [25.4K]

Answer:liquids and gases

Explanation:

friction

4 0

2 years ago

Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s

Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0

2 years ago

Encountering problems during separation/purification is a very common issue in organic synthesis, but these steps are crucial to

Rus_ich [418]

True is the correct answer! your welcome! :)

8 0

2 years ago