Why do lakes only freeze at the surface?

Does the pH increase or decrease, and does it do so to a large or small extent, with each of the following additions?(c) 5 drops

Umnica [9.8K]

The pH decreases to a large or small extent with each of the given additions.

<h3>What is common name of NaOH?</h3>

The common name of NaOH is sodium hydroxide. Lye and caustic soda are other names for sodium hydroxide, an inorganic substance having the formula NaOH. It is a white, solid ionic substance made up of the cations sodium (Na+) and the anions hydroxide (OH). Sodium hydroxide is a chemical that manufacturers utilize to make things like soap, rayon, paper, explosives, colors, and petroleum products. Processing cotton fabrics, metal cleaning and processing, oxide coating, electroplating, and electrolytic extraction are further uses for sodium hydroxide. A caustic metallic base is sodium hydroxide (NaOH), sometimes referred to as lye or caustic soda. Caustic soda, an alkali, is commonly employed in a variety of sectors, primarily as a potent chemical base in the production of pulp and paper, textiles, drinking water, and detergents.

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6 0

1 year ago

Which condition applies when water boils at 100oC

VLD [36.1K]

An ambient pressure of 1 atm

5 0

3 years ago

What two properties most affect the strength of the gravitational forces

goblinko [34]

1) Masses of the object

2) Distance between them

7 0

3 years ago

Read 2 more answers

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

Consider the following equilibrium: 4KO2(s) + 2H2O(g) 4KOH(s) + 3O2(g) Which of the following is a correct equilibrium expressio

jeka57 [31]

To be able to write correctly the equilibrium expression of a reaction, we need to know the balanced reaction and the phases of the substances in the reaction. When substances are solid, pure liquid they are not included in the expression. We do as follows:

<span>4KO2(s) + 2H2O(g) = 4KOH(s) + 3O2(g)

K = [O2]^3 / [H2O]^2</span>

5 0

3 years ago