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3 years ago

14

What do members of the same group have in common?

2 answers:

lord [1]3 years ago

7 0

The elements in each group have<span> the </span>same<span> number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements. Every element in the first column (</span>group<span> one) </span>has<span> one electron in its outer shell.</span>

Juli2301 [7.4K]3 years ago

6 0

Depends on what the group is but it normally means they share a common interest or goal.

You might be interested in

How do balanced chemical equations show the conservation of mass?

stepladder [879]

You cannot create something from nothing and similarly you cannot create less from more, if you didn't balance an equation that needs to be balanced, your product might end up equalling more or less than the total mass of the two elements combined. Which is impossible

3 0

4 years ago

when this chemical equation is correctly balanced what is the coefficient of the nal molecule I2+Na2S2O3+Nal+Na2S4O6

Kitty [74]

I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆

k(NaI)=2

4 0

3 years ago

Example for next four problems:

Lena [83]

Answer:

Compound formula : HCl

Element H

# atoms:1

Elements Cl

#atoms : 1

gram formula weight (g) = 36.46

4 0

3 years ago

Bleach contains the active ingredient NaClO. Analysis of bleach involves two sequential redox reactions: First, bleach is reacte

Alisiya [41]

Answer:

0,31%

Explanation:

For the reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0,043 L × 0,117 M of sodium tiosulfate = 5,031x10⁻³ moles of S₂O₃²⁻

5,031x10⁻³ moles of S₂O₃²⁻ × $\frac{1 I_2 mol}{2 S_{2}O_3 mol^{2-}}$ = <em>2,5156x10⁻³ moles of I₂</em>

These moles of I₂ were produced from:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2,5156x10⁻³ moles of I₂ ≡ moles of NaClO

2,5156x10⁻³ moles of NaClO × $\frac{74,44 g}{1mol}$ =<em> 0,187 g of NaClO</em>

Thus, percentage composition by mass is:

$\frac{0,187 g of NaClO}{60 g Of Bleach} x 100$ =<em> 0,31%</em>

I hope it helps!

5 0

3 years ago

A 1.25g sample of dry ice is added to a 765mL flask containing nitrogen gas at a temperature of 25.0°C and a pressure of 725 mmH

erica [24]

Answer : The total pressure in the flask is 1.86 atm.

Explanation :

First we have to calculate the pressure of $CO_2$ gas.

Using ideal gas equation :

$PV=nRT\\\\P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

where,

P = Pressure of $CO_2$ gas = ?

V = Volume of $CO_2$ gas = 765 mL = 0.765 L (1 L = 1000 mL)

n = number of moles

w = mass of $CO_2$ gas = 1.25 g

M = molar mass of $CO_2$ gas = 44 g/mol

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $CO_2$ gas = $25.0^oC=273+25.0=298K$

Putting values in above equation, we get:

$P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

$P_{CO_2}=\frac{1.25g}{44g/mol}\frac{(0.0821L.atm/mol.K)\times 298K}{0.765L}=0.909atm$

Now we have to calculate the total pressure in the flask.

$P_T=P_{N_2}+P_{CO_2}$

Given :

$P_{CO_2}=0.909atm$

$P_{N_2}=725mmHg=\frac{725}{760}=0.954atm$

conversion used : (1 atm = 760 mmHg)

Now put all the given values in the above expression, we get:

$P_T=0.954atm+0.909atm=1.86atm$

Therefore, the total pressure in the flask is 1.86 atm.

5 0

3 years ago