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Orlov [11]
3 years ago
5

When a Na atom loses an electron to a Cl atom, the two atoms have opposite charges and are attracted to one another. What is the

name of the bond that forms between these two atoms?

Chemistry
1 answer:
ahrayia [7]3 years ago
7 0

Answer:

Ionic bond

Explanation:

When Na loses an electron, it becomes positive ion.  

When Cl gains  an electron, it becomes negative ion.  

The electrostatic attraction between these ions is an ionic bond.

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What does the pacific tsunami warning system use to detect tsunamis ?
S_A_V [24]

Answer:

C

Explanation:

Their are pressure sensers in the water that will detect high pressure and set of scales that are currently detecting semic waves and trigger sierns.

7 0
3 years ago
0.28gram of NH3 on decomposition gave 0.25gram of nitrogen and hydrogen. Find the volume of hydrogen evolved at Ntp. (2gram hydr
hodyreva [135]

The volume of H₂ evolved at NTP=0.336 L

<h3>Further explanation</h3>

Reaction

Decomposition of NH₃

2NH₃ ⇒ N₂ + 3H₂

conservation mass : mass reactants=mass product

0.28 NH₃= 0.25 N₂ + 0.03 H₂

2 g H₂ = 22.4 L

so for 0.03 g :

\tt \dfrac{0.03}{2}\times 22.4=0.336

7 0
3 years ago
How many moles of gas are in a 1.0 liter canister if the temperature of the canister is 100 K and the pressure is 100 atmosphere
ladessa [460]

Answer:

n = 12.18 moles

Explanation:

Given that,

The volume of a canister, V = 1 L

The temperature of the canister, T = 100 K

Pressure, P = 100 atm

We need to find the number of moles of gas. Let there are n number of moles. We know that,

PV = nRT

Where

R is gas constant, R = 0.0821 L*atm/mol*K

n=\dfrac{PV}{RT}\\\\n=\dfrac{100\times 1}{0.0821 \times 100}\\\\n=12.18\ moles

Hence, there are 12.18 moles of gas.

4 0
2 years ago
PLEASE HELP ASAP! very much appreciated
soldier1979 [14.2K]

Answer:

-Unknown

Explanation:

Questions is not seen properly bro

3 0
2 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
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