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zlopas [31]
4 years ago
15

All of the following are balanced equations. State the

Chemistry
1 answer:
Advocard [28]4 years ago
7 0

Answer: a: reactants Na-2 Cl-2

Products: Na-2 Cl-2;

b: reactants P-1 Cl-13 H-6 Products P-1 H-6 Cl-13

c: reactants P-4 H-12 O-16

Products H-12 P-4 O-16

Explanation: since these equations are balanced the atoms on of element on the reactants side will be same as the atoms of the same element of the product side

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The difference in interchain stability between the polysaccharides glycogen and cellulose is due to: Group of answer choices the
denis-greek [22]

Answer: both the different glycosidic linkages of the molecules and the different hydrogen bonding partners of the individual chains.

Explanation:

Glycogen is a polysaccharide of glucose which is a form of energy storage in fungi, bacteria and animals. Glycogen is primarily stored in the liver cells and skeletal muscle.

The difference in interchain stability between the polysaccharides glycogen and cellulose is due to the different glycosidic linkages of the molecules and the different hydrogen bonding partners of the individual chains.

7 0
3 years ago
How many moles are there in 3.612 x 1024 molecules of CaO?
melamori03 [73]

The number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

<h3>Number of moles in the molecules</h3>

The number of moles in  3.612 x 10²⁴ molecules of CaO is calculated as follows;

6.02 x 10²³ molecules = 1 mole

3.612 x 10²⁴ molecules = ?

= (3.612 x 10²⁴ ) / (6.02 x 10²³ )

= 6 moles

Thus, the number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

Learn more about number of moles here: brainly.com/question/15356425

3 0
2 years ago
21 Which process is used to determine the concentration of an acid?
Inessa05 [86]

Answer:

4) Titration

Explanation:

Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.

8 0
4 years ago
Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL
Vladimir [108]

Answer:

<em><u>The molarity of the CoBr2•4H2O solution is  7.64 × 10-2 M</u></em>

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O  

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H  x 1.008 g/mol H = 8.064 g/mol

4 O  x 16.00 g/mol O = 64.00 g/mol

________________________________________

                           Sum = <u>290.79 g/ mo</u>

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x  \frac{ 1 mol  CoBr_2 . 4H_2O}{290.79 g CoBr_2 .  4H_2O}

= <u>0.0344  mol CoBr2•4H</u>

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344  mol CoBr2•4H  / 450 mL x 1 ml / 10^ -3 L

= 0.0764

=  7.64 × 10-2 mol/L

8 0
4 years ago
Formula semidesarrollada de los hidrocarburos aromáticos de 1 a 10 átomos de carbono
Tju [1.3M]

Answer:

El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:

El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.

Si se continúan colocando átomos de carbono con enlaces sencillos entre ellos e hidrógenos en los enlaces libres, se crean largas cadenas de compuestos. Al etano le sigue el propano (C2H8) y a éste, el butano (C4H10). Todos estos compuestos forman parte de la familia de los alcanos, y sus nombres terminan con el sufijo –ano para indicar que pertenecen a la misma familia.

3 0
3 years ago
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