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anygoal [31]
4 years ago
7

Two waves are traveling through the same container of air. Wave A has a wavelength of 2.0m. Wave B has a wavelength of 0.5m. The

speed of wave B must be ________ the speed of wave A.
Chemistry
2 answers:
evablogger [386]4 years ago
6 0
The answer is 2.0 m and that's it
mars1129 [50]4 years ago
4 0

Answer:

the same as

Explanation:

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Compare and Contrast the differences between plant animal cells.
sergij07 [2.7K]

Animal cells do not have a cell wall or chloroplasts but plant cells do. Animal cells are mostly round and irregular in shape while plant cells have fixed, rectangular shapes.

8 0
3 years ago
Lab Report
Diano4ka-milaya [45]

Answer:

The purpose of the experiment is to see how water of different temperature and salinity affect the density.

Explanation:

Temperature and salinity directly affect the density of the water. Water of low temperature is more dense than water of high temperature, BUT, (fresh)water with no salt is less dense than (sea)water with more salt, so temperature and salinity change density of water.

3 0
2 years ago
Need to know how to do chemical reactions
melomori [17]
Chemical reactions happen by them self like if you hear baking soda and it makes a gas.
8 0
3 years ago
A student determines that a 1.5 g mixture of CaCO3(s) and
zysi [14]

<u>We are given:</u>

Mass of the mixture = 1.5 grams

Number of Moles of NaHCO₃ = 0.01 mole

<u>Mass of 0.01 moles of NaHCO₃:</u>

Mass = number of moles * Molar mass

Mass = 0.01 * 84

Mass = 0.84 grams

<u>Mass% of Na in NaHCO₃:</u>

Molar mass of NaHCO₃ = 84 grams/mol

Molar mass of Na = 23 grams / mol

Mass% = (Molar mass of Na / Molar mass of NaHCO₃) * 100

Mass% = (23/84) * 100

Mass% = 27%

<u>Mass of Na in 0.01 moles:</u>

Since the Mass% of Na is 27% and the total mass is 0.84 grams

So, we can say that 27% of 0.84 is the mass of Na

Mass of Na = 0.84 * 0.27

Mass of Na = 0.23 grams

<u>Mass% of Na in the mixture:</u>

Total mass of the mixture = 1.5 grams

Mass of Na in the mixture = 0.23 grams

Mass% of of Na in the Mixture = (Mass of Na / Mass of Mixture) * 100

Mass% of Na = (0.23 / 1.5) * 100

Mass% of Na = 15.3%

Therefore, we have 15.3% Na in the given mixture by mass

7 0
3 years ago
Formula los siguientes compuesto: Dietil eter, Etanol, Propanotriol, Acido Propanodioico, Pentanal, Pentano-2,4-diona, Metanoato
9966 [12]

Answer:

Explanation:

En este caso para formular los compuestos, debes identificar el grupo funcional principal de la molecula. Una vez que eso está hecho, puedes intentar formularlo.

Empezaremos primero identificando el grupo funcional principal de la molécula, para luego formularlo correctamente.

Dietil eter: la terminación eter al final significa que pertenece al grupo de los éteres, el cual tiene como formula general R - O - R.

Etanol: debido a que termina en ol, este grupo pertenece a los alcoholes. Para formularlo solo se dibuja la molecula del etano, junto a un enlace con el grupo OH, como su formula general R - OH.

Propanotriol: igualmente termina en ol, por lo tanto es un alcohol, sin embargo, en este caso, tambien tiene la terminación prefija tri, asi que significa que hay 3 grupos OH en la molecula.

Acido propanodioico: esta es sencilla, porque empieza como acido, y solo hay un grupo funcional que empieza así y son los acidos carboxilicos, es decir, el grupo COOH (R - COOH) que es el carboxilo. Tiene el prefijo di, antes del oico, por lo que son dos carboxilos presentes en la molecula.

Pentanal: el sufijo al, significa que pertenece al grupo de los aldehidos, en este caso, posee el grupo carbonilo H - C = O.

Pentano - 2,4 - diona: la terminación ona significa que pertenece al grupo de las cetonas, (R - CO - R), parecido a los aldehidos, con la diferencia de que tiene grupos alquilos en lugar de un hidrogeno.

Metanoato de metilo: la terminación ato de ilo, pertenece a los esteres, (R - COOR) derivado de los acidos carboxilicos.

De aqui en adelante solo mencionaré los grupos funcionales pues ya se explicó el por que, por sus terminaciones:

Ciclohexano - 1.3 - diol: este pertenece a los alcoholes.

Acido heptanoico: acido carboxilico

Ciclobutil metil eter: eteres

Acetato de etilo: ester

2-metilbenzaldehído: aldehído unido a un grupo aromatico como el benceno.

Ciclohexanona: un ciclo (cadena cerrada) unido a un grupo carbonilo.

Butanona: cetona.

Observa la foto adjunta para que veas la formulación de cada una:

5 0
3 years ago
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