IUPAC name for BaSO3

La precipitación ácida es un agente de?

katrin2010 [14]

Answer:

b

Explanation:

por que si

6 0

3 years ago

Read 2 more answers

3. Joey pushes on a box with 15N of force to the left, and Annie pushes on the box with 18N of force to

mart [117]

Answer:

3N, Right

Explanation:

7 0

3 years ago

An object has a density of 1.75 g/mL. Will it float on water? Why?

Advocard [28]

Answer:

The object will not float in water because the density of the object is greater than that of water.

Explanation:

Density of object = 1.75 g/mL

Density of water = 1 g/mL

The Density of the object is 1.75 g/mL and that of water is 1 g/mL. This implies that the object is denser (i.e heavier) than water. Therefore, the object will not float in water, rather it will sink in water since it's density is greater than that of water.

4 0

4 years ago

A covalent bond between an oxygen atom and a hydrogen atom is best described as _____.

Igoryamba

The answer is most likely nonmetals. :)

7 0

3 years ago

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temp

Paul [167]

Solution: The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

If we have a hypothetical equation:

$A+2B\rightarrow 3C+5D$

Then the rate could be written as:

$rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}$

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

$rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}$

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of $O_2$ related to rate of consumption of $NH_3$ as:

$-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

multiply both sides by -5

$\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, rate of consumption of oxygen is $\frac{5}{4}$ the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

$\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 4

$\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}$

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of $H_2O$ to the rate of consumption of ammonia would be:

$\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 6

$\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, the rate of formation of $H_2O$ is $\frac{6}{4}$ times that is 1.5 times to the rate of consumption of ammonia.

5 0

4 years ago