All categories

4 years ago

A plant dying after being exposed to poison represents a physical change true or false?

Chemistry

2 answers:

Keith_Richards [23]4 years ago

6 0

False actually.

I'm sorry i don't know how i just got it right on my test and i hope you will?did to. And i hope i also get marked brainliest :)

r-ruslan [8.4K]4 years ago

3 0

Yes, i believe that it can have a physical change but it is most likely chemical.

You might be interested in

HELP PLZ!!!!!

Molodets [167]

The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

6 0

4 years ago

The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium

PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

$\frac{9 .84x10^8}{1000000}$ = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = $\frac{mass}{volume}$

mass = density x volume

mass = 1 x 984

= 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

4 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Imagine snow on top of a mountain. Describe at least two ways energy could be transferred as the seasons change?

Mademuasel [1]

Answer:

at summer season the snow will melt and will produce water and using generators we can produce electric energy and in spring and winter can produce electric energy also using the wind

8 0

2 years ago

Read 2 more answers

25 points please help

maxonik [38]

Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy ) × 100

For Accelerator 1 model,

% energy lost as waste = $\frac{663.1}{2078.3}\times100$ = 31.90%

For Accelerator 2 model,

% energy lost as waste = $\frac{2337.5}{7690.0}\times100$ = 30.39%

For Accelerator 3 model,

% energy lost as waste = $\frac{2259.6}{4061.9}\times100$ = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.

6 0

3 years ago