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marysya [2.9K]
3 years ago
15

What are cells and how are they important to maintaining life?

Chemistry
1 answer:
devlian [24]3 years ago
3 0

Answer:

Cells are the "building blocks for all living things." Without cells, we couldn't survive. Cells do many important things. Some of the things they can do are taking in food nutrients, frame our bodies, and change the food nutrients into energy.

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A solution was prepared by mixing 0.5000 m hno2 with 0.380 m no2-. ka = 4.58 x 10-4. calculate the ph of the solution. show work
adell [148]

pH of buffer can be calculated as:

pH=pKa+log[salt]/[Acid]

As ka = 4.58 x 10-4

Concentration of [Salt] that is NO2(-1)=0.380M

Concentration of [Acid] that is HNO2=0.500M

So, pH= -log(4.58*10^-4)+log((0.380)/0.500))

=3.21

So pH of solution will be 3.21

7 0
3 years ago
How many liters are present in 10 grams of H2O ?
Luba_88 [7]

Answer:

About 0.01

Explanation:

1 grams of H2O = 0.0010

0.0010 x 10 = 0.01

8 0
3 years ago
A student dissolves equal amounts of salt in equal amounts of warm water, room-temperature water, and ice water. Which result is
Tanya [424]

doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?

6 0
3 years ago
Read 2 more answers
A researcher wants to determine if a unicellular organism he discovered is an autotroph or a heterotroph. He radioactively label
GaryK [48]

Answer:

A

Explanation:

Autotrophs utilize the energy from  sunlight to reduce carbon dioxide to carbohydrates (glucose). The energy from the sunlight is used to split water into H+ and O2- and the H+ used in the reduction process. The labeled carbon in the carbon dioxide will, therefore, be incorporated by the autotrophs in the carbohydrates made in photosynthesis.  

4 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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