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Margaret [11]
2 years ago
14

During a long drought, many deer become weak because a basic need is not being met. What requirements for life are the deer most

likely lacking? enough space enough water adequate shelter ability to find mates
Chemistry
2 answers:
saw5 [17]2 years ago
3 0

Enough water, I actually took this test! <3

- S0ft1e

Veseljchak [2.6K]2 years ago
3 0

Answer: Water

Explanation: Drought=dry deer need water with no water to drink it’s easier for them to be hunted or killed

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Quien quiere ser mi amigo
vfiekz [6]

Answer:

Me encantaría

Explanation:

7 0
2 years ago
For question #12, use the following picture:<br><br> 12. What element is this? How do you know?
labwork [276]

Answer: The element shown in the image is Helium (He).

Explanation: We are given a image of an atom having protons, neutrons and electrons.

Number of protons as shown in image = 2

Number of neutrons as shown in image = 2

Number of electron as shown in image = 2

Atomic number = Number of protons = Number of electrons

Atomic number of the element = 2

Atomic Mass = Number of protons + Number of neutrons

Atomic mass = 2 + 2 = 4

The element having Atomic number = 2 and mass number = 4 is Helium.

Element = ^4_2\textrm{He}

8 0
3 years ago
Read 2 more answers
Help please! I'd appreciate it 
babymother [125]

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

6 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
2 years ago
Show your work. 37.2 moles CO2 to oxygen atoms.
Rufina [12.5K]
4 moles of oxygen (6.0zzx10

4 0
2 years ago
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