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3 years ago

Hi,Valency of chlorine is 1. Why?Thank you

Chemistry

2 answers:

dybincka [34]3 years ago

3 0

The person above me is correct

grin007 [14]3 years ago

3 0

Answer:

The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.

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How many representative particles are in one mole of CO2? PLEASE ANSWER ALMOST DUE

igor_vitrenko [27]

Answer:

I hope this helps, sorry if it doesnt. Not really good with chemistry, hope you have a great day/night :}

Explanation:

A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance.

12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

7 0

2 years ago

The process of cellular respiration causes a(n)

Galina-37 [17]

I would say 2 because co2 goes out and o goes in

4 0

1 year ago

Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)

GarryVolchara [31]

$\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1}$

Explanation

$\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)$

$\Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)
$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)

As an the most stable allotrope under standard conditions, $\Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}$

By definition,

$\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants})$

$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}}$

$\begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}$

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the $\Delta H\textdegree{} _f$ given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

$\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1}$ (Cox, Wagman, et al., 1984)

... yields $\Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}$ , which deviates significantly from the experimental value of $-94.76 \; \text{kJ}\cdot \text{mol}^{-1}$ (Chase, 1998.) It is thus assumed that the $\Delta H\textdegree{}_\text{rxn}$ value provided requires a reaction with $\text{I}_2 \; (s)$ rather than $\text{I}_2 \; (g)$ as a reactant.

3 0

3 years ago

What are the spectators ions in this reaction?

miv72 [106K]

Answer:

Option C. Na⁺(aq) and Cl¯(aq)

Explanation:

From the question given above, we obtained the following:

Na₂SO₄(aq) + CaCl₂(aq) —> CaSO₄(s) + 2NaCl(aq)

Ionic Equation:

2Na⁺(aq) + SO₄²¯(aq) + Ca²⁺(aq) + 2Cl¯(aq) —> CaSO₄(s) + 2Na⁺(aq) + 2Cl¯(aq)

From the ionic equation above, we can see that Na⁺(aq) and Cl¯(aq) are present on both side of the equation.

Therefore, Na⁺(aq) and Cl¯(aq) are the spectator ions because they did not participate directly in the reaction.

4 0

3 years ago

A 0.239 gram sample of a gas in a 100-mL flask exerts a pressure of 603 mm Hg at 14 degrees Celsius. What is the gas?

mamaluj [8]

Answer:

Explanation:

Given data:

Mass of gas = 0.239 g

Volume of gas = 100 mL

Pressure exerted by gas = 603 mmHg

Temperature of gas = 14 °C

What is gas = ?

Solution:

Volume of gas = 100 mL (100mL ×1 L/1000 mL= 0.1 L)

Pressure exerted by gas = 603 mmHg (603/760 = 0.79 atm)

Temperature of gas = 14°C ( 14+273 = 287 K)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

now we will calculate the number of moles.

n = PV/RT

n = 0.79 atm × 0.1 L / 0.0821 atm.L/ mol.K × 287 K

n = 0.079 /23.563 /mol

n = 0.003 mol

Molar mass of gas:

Number of moles = mass/molar mass

0.003 mol = 0.239 g/ molar mass

Molar mass = 0.239 g/ 0.003 mol

Molar mass = 79.7 g/mol

The molar mas of Br is 79.9 g/mol so it is closer to 79.7 thus given gas is Br.

6 0

3 years ago

Hi,Valency of chlorine is 1. Why?Thank you​

Hi,Valency of chlorine is 1. Why?Thank you