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2 years ago

When you see or visit grasslands, what do you know about the climate? (Select all that apply.)

Chemistry

2 answers:

denis23 [38]2 years ago

5 0

Answer:

They have little precipitation.

They have cold winters.

They have mild to hot summers.

Nana76 [90]2 years ago

4 0

When you see or visit grasslands you know that :

They have mild to hot summers ( B )
They have a lot of precipitation ( A )

<h3><u>Grassland vegetation</u> </h3>

Grasslands are a type of vegetation found mostly in areas with favourable weather conditions such as mild summers and right amount of precipitation which provides the grasslands with enough water for proper growth.

Other types of vegetation include :

Forest vegetation
Tundra and
Desert vegetation

Desert vegetation expreciences very hot summers and little precipitation which leads to the scarcity of green vegetation, while forest vegetation expreciences a very high level of precipitation.

Hence we can conclude that When you see or visit grasslands you know that :They have mild to hot summers ,They have a lot of precipitation

Learn more about grasslands : brainly.com/question/994161

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The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago

Igneous rocks would most likely be found near:

geniusboy [140]

Answer:

the deep sea floor. Known as the oceanic crust.

Explanation:

The deep seafloor (the oceanic crust) is made almost entirely of basaltic rocks, with peridotite underneath in the mantle. Basalts are also erupted above the Earth's great subduction zones, either in volcanic island arcs or along the edges of continents.

Hope this helps :)

7 0

3 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

6 0

3 years ago

3. Mr. Hill has 27 students in his class

stich3 [128]

Just dived both numbers by two

4 0

3 years ago

Read 2 more answers

Which solution has a molality of 0.25m nacl?

Solnce55 [7]

Molality can be expressed by moles of solute over kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl is equal to 0.25 moles of NaCl over 1 kg of water.

5 0

3 years ago

Read 2 more answers