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3 years ago

ASAP will give brainliest!!!!

Chemistry

1 answer:

Andru [333]3 years ago

7 0

Answer:

Explanation:

fluorine's atomic number is 9

electronic configuration: 2,7

so it needs 1 electron to stabilise

that's why 1 covalent bond

hope it helps!!

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The observed rotation of 2.0 gg of a compound in 50 mLmL of solution in a polarimeter tube 25-cmcm long is +47.2 ∘∘ . What is th

Anastaziya [24]

Answer:

[∝] = +472

Explanation:

Specific rotation in a solution is defined as:

[∝] = ∝ / c×l

Where:

[∝] is specific rotation, ∝ is observed rotation (In degrees), c is concentration in g/mL and l is path length (In dm).

∝: +47.2°

c: 2.0g / 50mL = 0.04g/mL

l: 25cm × (1dm /10cm) = 2.5dm

Replacing:

[∝] = +47.2° / 0.04g/mL×2.5dm = <em>+472</em>

I hope it helps!

4 0

3 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

How many carbon atoms are represented in the formula for CaMg3(CO3)4

Galina-37 [17]

The formula is for 1 Calcium, 3 Magnesium, 4 Carbon, & 12 Oxygen. The 4 after the parenthesis of (CO3) distributes to each element.

Answer : There are 4 Carbon atoms in the formula.

3 0

2 years ago

Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks

charle [14.2K]

Answer:

pH = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

4 0

3 years ago

How many moles of Na are needed to produce moles of NaCl in the reaction below

MArishka [77]

I think it’s 1 mole
If Iam not mistaken

3 0

3 years ago