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Ksivusya [100]
2 years ago
13

Match the part of the wave to the diagram

Chemistry
1 answer:
erik [133]2 years ago
5 0

Answer:

The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.

Explanation:

The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.

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The chart indicates the time, speed, and velocity of five runners.
Vitek1552 [10]

Answer:

garreth is the fasstis runner

luigi is the 2 fasts runner

elisa is the 3 fastes runner

Explanation:

3 0
3 years ago
A molecule of an organic compound contains at least one atom of
Mila [183]
A molecule of an organic compound contains at least one atom of "<span>(1) carbon". This is referred to as organic chemistry, since all living matter is carbon-based. </span>
5 0
3 years ago
Read 2 more answers
Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
mr_godi [17]

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
4 0
3 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ΔH° for the r
PolarNik [594]

Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]

\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]

\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]

Delta H=-105kJ

Therefore, the enthalpy change for this reaction is, -105 kJ

8 0
3 years ago
Why do Man holes(road) build by use only circular type??
sashaice [31]

Answer: Manholes are usually round, primarily because roundness is the best shape to resist the compression of the earth; covers are round because they are easier to manufacture than square or rectangular shapes, they are easier to move by rolling, and they can't fall into the opening.

Explanation:

7 0
2 years ago
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