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natulia [17]
3 years ago
14

which factor would increase the carrying capacity for a white-tailed deer in an eastern temperate forest ecosystem

Chemistry
2 answers:
ycow [4]3 years ago
6 0

Answer:

<u>Development of natural reserves of forest land and crack down on hunting</u>

Explanation:

  • White-tailed deer also known as Virginia deer is a small to medium-sized deer common to North America, and South America. Deer coat is reddish-brown in spring and summer they turn grey-brown through the fall in winter.
  • A resident of the temperate grasslands adapted themselves to open lands of prairie, and savanna woodlands. They commonly eat leguminous crops and forge on other crops including leaves, cacti, and shoots. They also eat acorns, fruit, and corn.  
  • Due to climate change and the influence of man majority of deer have migrated and the impact of diseases has also reduced the large deep population which can be overcome by rehabilitation programs so as their original habitats are ensured.
saveliy_v [14]3 years ago
4 0
This is a guess, but I think it could be the lack of limiting factors.
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An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated
pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

  • 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
  • n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

  • 2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

  • C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
  • H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
  • N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0
3 years ago
Write the products of photosystem i and photosystem ii
Margaret [11]
<span>ATP,O2 and NADPH are the </span>products<span>. H2O,NADP,ADP and Pi are the reactants. acts as an electron carrier between the cytochrome b6f and </span>photosystem 1 (PS1) complexes in the photosynthetic electron-transfer chain.

Photosystem II<span> (or water-plastoquinone oxidoreductase) is the first protein complex in the light-dependent reactions of oxygenic photosynthesis. It is located in the thylakoid membrane of plants, algae, and cyanobacteria.</span>
3 0
3 years ago
Read 2 more answers
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
For 30 points! Please help me understand this question
oksian1 [2.3K]

Answer:

D) Adding a catalyst

Explanation:

Adding a catalyst decreases activation energy and allows the reaction to occur more easily.

7 0
3 years ago
A football field is 120 yards by 53.333 yards. What is the area of the football field in acres if 1 acre=43560 ft?? Use correct
maw [93]

Answer:

1.3223 acres

Explanation:

a football field's area is 360 feet (120 yards) x 160 feet (53.333 yards) = 57,600 sq. feet

if the area of an acre is 43,560 sq. feet, then a football field in acres = 57,600 sq. feet / 43,560 sq. feet = 1.3223 acres

we can verify our answer by doing the same calculation in sq. yards:

football field = 120 yards x 53.33 yards = 6,400 sq. yards

an acre is 4,840 sq. yards

football field in acres = 6,400 sq. yards / 4,840 sq. yards = 1.3223 acres

8 0
3 years ago
Read 2 more answers
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