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VARVARA [1.3K]
3 years ago
7

Why doesn't grassland biome naturally support trees

Chemistry
1 answer:
ra1l [238]3 years ago
4 0
Each different species of grass grows best in a particular grassland environment (determined by temperature, rainfall, and soil conditions). The seasonal drought, occasional fires, and grazing by large mammals all prevent woody shrubs and trees from invading and becoming established.
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What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?
Step2247 [10]

Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL

Best regards.

3 0
3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
In a chemical reaction between copper metal and silver nitrate, 12.7 g of copper(II) and 64.5 g of silver nitrate react, and 38.
hodyreva [135]

Answer:What should you do if you realize during research that your original theory is wrong

Explanation:

8 0
3 years ago
Hsvsusvshssveuevsjsvsjdvsudvysvwkwhsjwvsgsbkwbsywvejwvsh​
dybincka [34]

Sorry, I won't understand your words.

6 0
3 years ago
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The product of alpha decay of the Nobelium isotope 259No is
Svet_ta [14]

Answer:

Nobelium is made by the bombardment of curium (Cm) with carbon nuclei. Its most stable isotope, 259No, has a half-life of 58 minutes and decays to Fermium (255Fm) through alpha decay or to Mendelevium (259Md) through electron capture.

Explanation:

7 0
3 years ago
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