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qwelly [4]
3 years ago
5

Which organisms are secondary consumers in a temperate coniferous forest? (Select all that apply.)

Chemistry
2 answers:
brilliants [131]3 years ago
8 0

Answer:

Lynx and wolf

Explanation:

Ira Lisetskai [31]3 years ago
6 0
Lynx and wolf!!! (: hope this helps
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One peanut M&M weighs approximately 2.33 g.
Sphinxa [80]

Answer:

There are 23076 peanut M&M's in 53.768 kg of M&M's.

Explanation:

First we <u>convert 53.768 kg into g</u>:

  • 53.768 kg * 1000 = 53768 g

Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:

  • 53768 g / 2.33 g = 23076

So there are 23076 peanut M&M's in 53.768 kg of M&M's.

3 0
2 years ago
When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th
maxonik [38]
I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture. 

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>

8 0
3 years ago
What is the threshold frequency ν0 of cesium?
Montano1993 [528]
 I think this is what you're after:

 Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1] 


Convert to J and divide by the Avogadro Const to give E in J per photon 


E = 375700/6.022×10^23 = 6.239×10^-19 J 


Plank relationship E = h×ν E in J ν = frequency (Hz s-1) 


Planck constant h = 6.626×10^-34 J s 


6.239×10^-19 = (6.626×10^-34)ν 


ν = 9.42×10^14 s^-1 (Hz) 


IP are usually given in ev Cs 3.894 eV 


<span>E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon </span>

4 0
2 years ago
Embryos have the genes of
REY [17]

Answer:

gggggggg

Explanation:

4 0
3 years ago
Read 2 more answers
GIVING BRAINLIEST!! PLEASE HELP
LenaWriter [7]

Answer:

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Explanation:

8 0
3 years ago
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