What results when two or more atoms share electrons equally?

vovangra [49]

Answer:

1. an object that has been intentionally placed into orbit. they are called Artificial Satalites.Satanists.

2. a huge collection of gas ,dust and billions of stars and their solar systems held together by gravity.

3. a large object such as Jupiter or Earth that orbits a star. planets are smaller than stars and they do not produce light.

4. July 16, 1969

7 0

3 years ago

Chemical formula for tin iv carbide

aniked [119]

Answer:

Explanation:

Sn(WC)2

if it is tungsten carbide this should be correct but there are many versions of carbide

Sn(MC2)2

could also be possible

the 2 next to MC should be a subscript

3 0

3 years ago

La columna de la izquierda corresponde a los tipos de sales y la columna derecha a los tipos de fórmula que presentan. Relaciona

Galina-37 [17]

Answer:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas b. MXO

4. Basicas d. M(OH)X

Explanation:

¡Hola!

En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.

Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas b. MXO

4. Basicas d. M(OH)XO

En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.

¡Saludos!

3 0

2 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago

The equilibrium constant for the reaction of carbon monoxide with water is 1.845. if 1.00 mol of each reactant is placed in a 2.

SpyIntel [72]

[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

initial 0.5 0.5 0 0

change -X -X +X +X

Equ (0.5-X) (0.5-X) X X

when Kc = X^2 * (0.5-X)^2

by substitution:

1.845 = X^2 * (0.5-X)^2 by solving for X

∴X = 0.26

∴ [CO2] = X = 0.26

4 0

2 years ago