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2 years ago

The halogens are in Group 7 of the periodic table.

Chemistry

1 answer:

Kaylis [27]2 years ago

6 0

Answer:

Explanation:

When it comes to the trend of reactivity in halogens, the reactivity decreases down the table. This is because in the nucleus of the atoms (mix of protons and neutrons), there is less of an attraction of the protons to the electrons on the outer shell, which gets further away going down. Since halogens are non metals, they need to gain electrons to become stable, and so this lesser attraction to the electrons may not allow this at least immediately. Fluorine is the most reactive out of all, have less of its outer shells and so being able to react much quickly.

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1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

3 years ago

A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base

Whitepunk [10]

Answer:

at the beginning:

pH = 0.745

Explanation:

HCl is a strong acid, so:

HCl + H2O → H3O+ + Cl-

0.18 M 0.18 0.18.....equilibrium

before base is added:

∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )

⇒ pH = 0.745

8 0

3 years ago

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andrezito [222]

Explanation:

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3 years ago

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Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneo

Mashutka [201]

Answer:

a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance

Explanation:

Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.

8 0

3 years ago