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user100 [1]
3 years ago
7

Draw a Lewis structure for C2H3Cl . Include all hydrogen atoms and show all unshared electron pairs. None of the atoms bears a f

ormal charge, and all atoms have octets (except for hydrogen atoms, which have duets).
Chemistry
1 answer:
Korvikt [17]3 years ago
5 0

Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

Best regards!

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A- the 3 is along w the PO4, which would make it part of that bond

C- there is no 3 at all for Na in this choice, making it incorrect

D- Again, the 3 is placed on the other half of the bond

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Which of the following summarizes photosynthesis?
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Carbon dioxide + water → glucose + oxygen + water
6 0
4 years ago
Se trata 50 g de aluminio que posee 10% de impurezas con suficiente cantidad de ácido sulfúrico ¿qué peso de sulfato de aluminio
Bogdan [553]

Respuesta:

199.5 g

Explicación:

Paso 1: Escribir la reacción balanceada

2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂

Paso 2: Calcular la masa pura de 50 g  de Al

Aluminio tiene 10% de impurezas, es decir, 10% de 50 g = 5 g. Luego, tiene 50 g - 5 g = 45 g de Al puro.

Paso 3: Calcular la masa teórica de Al₂(SO₄)₃ obtenida a partir de 45 g de Al

La relación de masas de Al₂(SO₄)₃ a Al es 342:54.

45 g Al × 342 g Al₂(SO₄)₃/54 g Al = 285 g Al₂(SO₄)₃

Paso 4: Calcular la masa real de Al₂(SO₄)₃ obtenida

El rendimiento de la reacción es de 70%.

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4 0
3 years ago
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe:
sladkih [1.3K]

Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

= \frac{Mass}{Molar\ mass}

= \frac{10.0 g}{259.65g / mol}

= 0.0385 mol

Mass of PbO needed is

= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

7 0
3 years ago
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