H2(g) + I2(g) → 2HI(g) Which of the following will decrease the rate of the forward reaction? A. add some H2 B. remove some HI C. increase the volume D. increase the temperature

h2 {g} would go into the fug so then u tage the oother rate then put the temp so it would be d

Explanation:

130 ppm Pb

Explanation:

We use parts per million to express the concentrations of solutions that contain very, very small amounts, often called trace amounts, of a given solute.

More specifically, a solution's concentration in parts per millions tells you the number of parts of solute present for every

10

6

=

1

,

000

,

000

parts of solution. You can thus say that a

1 ppm

solution will contain exactly

1 g

of solute for every

10

6

g

of solution.

In your case, you know that you have

38

mg Pb

⋅

1 g

10

3

mg

=

3.8

⋅

10

−

2

.

g Pb

in exactly

300.0 g

=

3.000

⋅

10

2

.

g solution

This means that you can use this known composition as a conversion factor to scale up the mass of the solution to

10

6

g

10

6

g solution

⋅

3.8

⋅

10

−

2

.

g Pb

3.000

⋅

10

2

g solution

=

130 g Pb

Since this represents the mass of lead present in exactly

10

6

g

of solution, you can say that the solution has a concentration of

concentration

ppm

=

130 ppm Pb

−−−−−−−−−−−−−−−−−−−−−−−−−−−−

The answer is rounded to two sig figs, the number of sig figs you have for the mass of lead present in the sample.