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Flura [38]
3 years ago
8

Eugene describes the physical property of a material as “sweet and floral.” What physical property of the material is Eugene mos

t likely observing?
boiling point
shape
hardness
odor
Chemistry
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

D

Explanation:

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Indicate whether the following represents a Chemical or Physical change: Milk sours
marusya05 [52]

Answer:

Chemical Change

Explanation:

Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.

4 0
3 years ago
Please answer correctly ...I will mark you brainliest
Valentin [98]

Answer:

Ali will need to either taste test or test the drinks on a ph scale. A neutral drink is water, as neutral has a ph of 7. The ph scale ranges from 1 to 14, going from acidic to basic. So if the drink has a ph less than 7, it is acidic, if its more, its basic.

4 0
3 years ago
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The atomic number of magnesium (mg) is 12. this is also the number of which of the following
Rzqust [24]
The answer is protons.
8 0
3 years ago
Stainless steel is an example of a ____ solution
Margaret [11]
I believe the correct answer from the choices listed above is option B. Stainless steel is an example of a solid-solid solution. It is an alloy which is made up of different metals <span>such as </span>carbon<span>, </span>manganese<span>, phosphorus, sulfur, nickel, chromium and others. Hope this answers the question.</span>
5 0
3 years ago
4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate
Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0
3 years ago
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