how much heat is absorbed when 15.0 g of liquid water is heated until the temperature increases by 16.0.

Why Are Microorganism's Used to separate waste products from water?

vampirchik [111]

Answer:

Untreated water from rivers, lakes, ponds, springs, or streams is likely to contain unsafe levels of infectious microorganisms, and drinking it may cause illness.

Explanation:

6 0

2 years ago

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$

Finally, we can put the ka value and solve for "X":

$2.2X10^-^5=\frac{X*X}{0.004-X}$

$2.2X10^-^5=\frac{X^2}{0.004-X}$

$X= 0.000285$

So, we have a concentration of 0.000285 for $H^+$ . With this in mind, we can calculate the pH value:

$pH=-Log[H^+]=-Log[0.000285]=3.55$

I hope it helps!

8 0

2 years ago

Water is an example of a

d1i1m1o1n [39]

Water is an example of a compound. It is a compound because it is made up of two hydrogen atoms and one oxygen atom hence its formula H2O.

8 0

2 years ago

You will be putting a drop of hydrogen peroxide on the edge of your plate as one of many tests used to identify bacteria. If it

irga5000 [103]

Answer:

I'd say ''B, positive reaction''

Hope this helps ^-^* Sorry if I'm Wrong.

Explanation:

8 0

2 years ago

Given the equation representing a reaction:

Lesechka [4]

Oxidation-reactions are equations where some species release electrons and other species gain electron.

So two reactions happen simultaneously the oxidation of one species and the reduction off the other species.

In the oxidations, the species release electrons and, consequently, increase their oxidation number.

In the reductions, the species gain electrons and, consequently, decrease their oxidation number.

This last reaction, reduction, is what the above equation represents: Sn 4+ is gaining 2 electrons and so its oxidation number decrease from 4+ to 2+.

In, conclusion that is a reduction equation (option d)

7 0

3 years ago