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Marina86 [1]
3 years ago
11

You are riding a bicycle. If you apply a forward force of 125 N, and you and

Chemistry
2 answers:
Slav-nsk [51]3 years ago
6 0

Answer:

1.25m/s^2

Explanation:

I tried what the first dude said and got it wrong

GenaCL600 [577]3 years ago
4 0

Answer:

its d

Explanation:

just did it on apex

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Describe the structure of an Atom (HELP!)
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Answer:

The answer is A.

Explanation:

Neutrons and protons are located in the dense middle of the atom called the nucleus, and electrons are located on the electron cloud located outside of the nucleus.

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3 years ago
Water and oxygen gas are the products of a chemical reaction.
Darina [25.2K]
The answer is C.

H₂O₂ ----> H₂O + O₂
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The pattern of inheritance in which more than one pair of genes affects a trait is *
s344n2d4d5 [400]

Answer:

hello, the answer is polygenic:)

8 0
3 years ago
Calculate the energy of a photon of 765 nm wavelength light in J.
weqwewe [10]

Answer:

2.6×10^-19 J/photon

Explanation:

E  of photon  =  h  ×  ν

where h= 6.63 × 10^-34 j.s

v= C ÷ λ

E = ( h × c) ÷ λ

E = (6.63 × 10^-34 × 3.00 ×10^8 )  ÷ ( 765 × 10^-9)

E = 2.6×10^-19 J/photon

3 0
3 years ago
En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
3 years ago
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