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7nadin3 [17]
3 years ago
13

This exercise deals with safety issues in the Freezing Point Depression experiment. Choose the answer that best completes each s

tatement. (a) Stearic and lauric acids are 1 but prolonged contact with skin may cause 2 . (b) If either stearic acid or lauric acid is spilled on the skin, it should be
Chemistry
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

A1 - non hazardous

A2 - irritation

B. Washed off with soap and water

Explanation:

- all materials or chemicals that are not specifically deemed hazardous that is does not cause any harm to the organism in the presence of the material or chemical.

- irritation is a discomfort to an organism which can be caused by an external substance like chemicals.

B. This is the action carried out in order to clean a substance off a surface of a body.

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6 0
3 years ago
Which diagram shows a pair of electrons that have opposite spins?
Nastasia [14]

It's B ------------------------------------

7 0
3 years ago
Read 2 more answers
A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t
Margarita [4]

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

7 0
3 years ago
I need help on this please
tino4ka555 [31]

Answer:

The answer is c

Explanation:c

3 0
3 years ago
I NEED THE ANSWER IMMEDIATELY!!! Pls helppp
Wewaii [24]

Answer:

turgor pressure can be done in a lab or a self test.

turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.

Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.

Explanation:

  • salt is bad for turgor pressure.
  • Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
  • The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
7 0
3 years ago
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