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steposvetlana [31]
3 years ago
10

In order to survive, plants absorb sunlight, water, and nutrients from the soil. What property is this?

Chemistry
1 answer:
zvonat [6]3 years ago
4 0

Plants are unique organisms that can absorb nutrients and water through their root system, as well as carbon dioxide from the atmosphere. Soil quality and climate are the major determinants of plant distribution and growth.

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Which is an example of the strong force in action?
yarga [219]

I would say that it would probably be: The Sun pulls on Earth as Earth moves around the sun.

Sorry if I'm wrong <3

3 0
2 years ago
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Is rubbing alcohol polar, non-polar, or ionic?
Karolina [17]

Answer: Rubbing alcohol molecules have a polar and nonpolar part, which means they are able to form hydrogen bonds with water and therefore able to mix with it.

Explanation:

7 0
2 years ago
Why is it that the decomposition of nitroglycerin can result in so
patriot [66]

Answer:

with the molecular formula C3H5(ONO2)3, has a high nitrogen content (18.5 percent) and contains sufficient oxygen atoms to oxidize the carbon and hydrogen atoms while nitrogen is being liberated, so that it is one of the most powerful explosives known.

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3 0
3 years ago
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol
Dafna11 [192]

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

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15.1 mol × 44.01 g/mol = 665 g

4 0
3 years ago
You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.
Nitella [24]

The question requires us to use the dilution formula M_iV_i=M_fV_f, where M_i and V_i are the stock concentration and volume respectively, then M_f and V_f are the dilute concentration and volume respectively.

a. C_s_t_o_c_k= 1.5 M KCI, C_d_i_l_u_t_e=0.20M KCl, V_d_i_l_u_t_e=250ml KCl

b.M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.


8 0
3 years ago
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