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levacccp [35]
3 years ago
11

Food couldn’t reach the stomach without the esophagus and the _? please help only correct answers.

Chemistry
1 answer:
expeople1 [14]3 years ago
3 0
<span>Food couldn’t reach the stomach without the esophagus and the throat.</span>
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4. Juan tiene un cilindro de 40 libras herméticamente cerrado con gas propano y lo conecta mediante una válvula a otro cilindro
ch4aika [34]

Answer:

Ver explicacion

Explanation:

Cuando el cilindro de 40 libras está conectado al cilindro de 100 libras, generamos una presión que continúa hasta que los dos cilindros alcanzan la misma presión de gas en equilibrio. Recuerde que el cilindro de 100 libras estaba inicialmente vacío. Esto significa que su presión inicial es 0. El cilindro de 40 libras ya estaba lleno, por lo que dividimos esta cantidad en dos para tener en cuenta su distribución entre los dos cilindros.

Ahora tenemos 20 libras de gas propano presentes en cada cilindro. La implicación de esto es que, en el cilindro de 100 libras, necesitamos 80 libras adicionales para completar las 100 libras.

5 0
3 years ago
Need help with chemistry
AlekseyPX

Answer:

you need to use the 2 because I already did it

Explanation:

db

3 0
2 years ago
What do the dots around a symbol in a Lewis structure represent?
anzhelika [568]
B. valence electrons
3 0
3 years ago
what is known as electronegativity and mention how the electro negativity vary long the periods and down the groups..please i ne
Kobotan [32]

Answer:

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

5 0
3 years ago
Read 2 more answers
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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