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3 years ago

Examine the model of a metallic bond.

Chemistry

1 answer:

Sindrei [870]3 years ago

3 0

Answer:

b the delocalized

Explanation:

it could be the right answer

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What is the answers?

ra1l [238]

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

5 0

3 years ago

Pls someone help me out

SpyIntel [72]

Do you still need help

4 0

2 years ago

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

A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution

madam [21]

<u>Answer:</u> The initial pH of the HCl solution is 3

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=100mL\\n_2=1\\M_2=0.10M\\V_2=1mL$

Putting values in above equation, we get:

$1\times M_1\times 100=1\times 0.10\times 1\\\\M_1=\frac{1\times 0.10\times 1}{1\times 100}=10M$

1 mole of HCl produces 1 mole of $H^+$ ions and 1 mole of $Cl^-$ ions

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.001M$

Putting values in above equation, we get:

$pH=-\log (0.001)\\\\pH=3$

Hence, the initial pH of the HCl solution is 3

6 0

3 years ago

Morgan wants to compare the weather conditions at her school each day during the week. She decides to measure and record weather

Firlakuza [10]

Answer: Air Temperature

8 0

3 years ago