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3 years ago

Which of the following particles is the smallest in radius

Chemistry

2 answers:

Gnesinka [82]3 years ago

7 0

Answer:

Its helium

Explanation: Because teh effective nuclear charge holds the valence ectrons clos to the nuclues. the Atomic radius would decrease as you move from the led=ft and right of the perodic table

rusak2 [61]3 years ago

5 0

Helium is the smallest radius

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Children tend to resemble their parents in many ways; they may have like eye color, hair texture, height, or skin tone. This is

weeeeeb [17]

black would be BB .

dominant gene is B.

phenotyope is where it says white fur.

recessive gene is b.

white is bb

5 0

3 years ago

How many milliseconds (ms) are there in 3.5 seconds (s)?

ivolga24 [154]

Answer:

13 hundred

Explanation:

4 0

3 years ago

Read 2 more answers

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0

3 years ago

Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K

yawa3891 [41]

Answer:

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

Explanation:

$Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)$

Concentration of nickel ion = $[Ni^{2+}]=x$

Concentration of nickel complex= $[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L$

Concentration of ethylenediamine = $[en]=\frac{0.80 mol}{2 L}=0.40 mol/L$

The formation constant of the complex = $K_f=4.0\times 10^{18}$

The expression of formation constant is given as:

$K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}$

$4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}$

$x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}$

$x=3.125\times 10^{-19} mol/L$

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

6 0

4 years ago