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NemiM [27]
3 years ago
14

Hey y’all, I’m growing a plant for science class but I need some questions about the plant, can someone please give me atleast 3

questions to answer for it.. can you give me atleast one tricky one
Hurry pls <3
Chemistry
1 answer:
Anna35 [415]3 years ago
3 0

Answer:

What type of plant

Where is it mainly found

What are some facts about it?

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The Resource Conservation and Recovery Act controls hazardous waste from its creation to its disposal. Please select the best an
o-na [289]

Answer:

True

Explanation:

Resource Conservation and Recovery Act (RCRA). 1976 of United States Environmental Protection Agency(EPA) empowers EPA to control the production, transportation, storage, treatment and disposal of hazardous waste. The RCRA act was amended in 1984 and 1986 to include Waste minimization along with appropriate disposal (not in the landfill site) and tackling of petroleum hazardous waste respectively along with other waste.

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3 years ago
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What is liquor ammonia fortis?​
erik [133]

Answer:

Ammonia fortis liquor is a saturated solution of ammonia in water. It is also called 880 ammonia. Its relative density is 0.880. It is stored in tightly sealed bottles in a cold place. (Sorry if I'm wrong)

Explanation:

6 0
3 years ago
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What is used up in and stops a chemical reaction?
CaHeK987 [17]
B - limiting reactant
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3 years ago
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A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp
frozen [14]

Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y  = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y

Replacing z = 0.25; we have:

0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y

replacing z = 0.25

0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y

0.75 = x + 2y

6 0
2 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
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