What is the difference between a stable and unstable isotope

Water, H2O, is a molecule made of oxygen and hydrogen. The bonds that hold water molecules together are due to shared __________

adell [148]

Oxygen and hydrogen share electrons in the molecule of water to form covalent bonds.

<h3>What kinds of bonds exist?</h3>

Covalent bonds: These are formed between nonmetals and electrons by sharing electrons.
Ionic bonds: These are formed between metals, which lose electrons, and nonmetals, which gain electrons.
Metallic bonds: There are formed between metals. Electrons are delocalized in a cloud.

Water, H₂O, is a molecule made of 2 nonmetals: oxygen and hydrogen. The bonds that hold water molecules together are due to shared electrons, and known as covalent bonds.

Oxygen and hydrogen share electrons in the molecule of water to form covalent bonds.

Learn more about chemical bonds here: brainly.com/question/6071754

5 0

2 years ago

at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were

Nataly [62]

<u>Answer:</u> The percent change in volume will be 25 %

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1$

Putting values in above equation, we get:

$\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L$

Percent change of volume = $\frac{\text{Change in volume}}{\text{Initial volume}}\times 100$

Percent change of volume = $\frac{(2-1.5)}{2}\times 100=25\%$

Hence, the percent change in volume will be 25 %

5 0

2 years ago

write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi

diamong [38]

Answer:

First ionization of lithium:

$\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}$ .

Second ionization of lithium:

$\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}$ .

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol $(g)$ next to any atom or ion in the equation.
The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

The products shall contain a gaseous ion with charge +1 $\text{Li}^{+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: $\text{Li}\;(g)$ .

Hence the equation: $\text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}$ .

Second Ionization Energy of Li:

The product shall contain a gaseous ion with charge +2: $\text{Li}^{2+}\;(g)$ as well as an electron $\text{e}^{-}$ .
Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?

The equation for this process is $\text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}$ .

5 0

2 years ago

A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If t

11Alexandr11 [23.1K]

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

8 0

2 years ago

Read 2 more answers

how many moles of sodium hydroxide would have to be added to 250 ml of a 0.403 m acetic acid solution, in order to prepare a buf

jonny [76]

The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.

Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.

pH=pKa + log [salt]/[acid]

Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid and 250 ml is added in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.

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6 0

6 months ago