How does the mass of an oxygen nucleus

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. Which of these compounds will precipitate out of this combined s

Archy [21]

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. BaS0₄ (s) will precipitate out of this combined solution.

Molecular equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 KNO3(aq)?

This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:

A-B + C-D → A-D + C-B;

where A and C are cations, and B and D are anions.

Complete ionic equation: Includes all ions and the precipitate.

2K^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2[NO3]^-(aq) → 2K^+(aq) + 2[NO3]^- + BaSO4(s)

In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.

The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.

Learn more about precipitation here : brainly.com/question/1783904

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6 0

1 year ago

Rank the ions in each set in order of decreasing size, and explain your ranking:

Valentin [98]

Answer:

a) Se²⁻> S²⁻ > O²

b) Te²⁻ > I- >Cs+

c) Cs+ > Ba²⁺ > Sr²⁺

Explanation:

(a) Se²⁻, S²⁻, O²⁻

In general, ionic radius decreases with increasing positive charge.

As the charge on the ion becomes more positive, there are fewer electrons.

The ion has a smaller radius. In general, ionic radius increases with increasing negative charge.

For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table

Se²⁻> S²⁻ > O²

(b) Te²⁻, Cs⁺, I⁻

Te²⁻ > I- >Cs+

Te2- hast the biggest size, because of the double negative charge.

Cs+ has the smallest size since it has the most positive charge, compared to Te2- and I-.

(c) Sr²⁺, Ba²⁺, Cs⁺

Cs+ > Ba²⁺ > Sr²⁺

Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.

Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+)

5 0

3 years ago

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

diamong [38]

Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

$G = \frac{0.02 * 40.04}{ 100}$

$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

6 0

3 years ago

What are isotopes..??

tatuchka [14]

Answer:

each of two or more forms of the same element that contain equal numbers of protons but different number of nurtrons

in there nucleir

Explanation:

.....

6 0

3 years ago

Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$