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3 years ago

5

For the main group elements( excluding the transition elements) , is it necessary to memorize the type of ion each element makes

or could you predict the ion charge using a periodic table?

1 answer:

ella [17]3 years ago

7 0

You can predict it based of the electronegativity

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If you had a 5 gram sample of lawrencium how much would still remain in 30 minutes

mart [117]

Answer:

$\large \boxed{\text{4.5 g}}$

Explanation:

There are eight isotopes of lawrencium, and they are all radioactive,

I will do the calculation for ²⁶²Lr, which has a half-life of 3.6 h.

Let A₀ = the original amount of lawrencium.

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀, and so on.

We can write a general formula for the amount remaining:

A =A₀(½)ⁿ

where n is the number of half-lives

$n = \dfrac{t}{t_{\frac{1}{2}}}$

Data:

A₀ = 5 g

t = 30 min

$t_{\frac{1}{2}} = \text{3.6 h}$

Calculations:

(a) Convert the half-life to minutes

$t_{\frac{1}{2} }= \text{3.6 h} \times \dfrac{\text{60 min}}{\text{1 h}} = \text{216 min}$

(b) Calculate n

$n = \dfrac{30}{216} = 0.1389$

(c) Calculate A

$A = \text{5 g} \times \left (\dfrac{1}{2}\right)^{0.1389} = \text{5 g} \times 0.9082 = \textbf{4.5 g}\\\\\text{The mass of lawrencium remaining after 30 min is $\large \boxed{\textbf{4.5 g}}$}$

8 0

4 years ago

Balanced equation for the equation between zinc and acetic acid

eduard

Zn(s) + 2CH3COOH(aq) → (CH3COO)2Zn(aq) + H2(g)<span>
</span>

5 0

3 years ago

For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained f

N76 [4]

Answer:

The dynamic viscosity of the liquid is 0.727 kg/m*s

Explanation:

In the equation for that viscosimeter, ν = KR⁴t, <u>the terms K and R are not dependent on the liquid that is being tested</u>, unlike ν and t.

Using that equation and the data given in the problem, we can calculate the product of K and R⁴.

1.19*10⁻³m²/s = (KR⁴)* 1430 s

KR⁴=8,32*10⁻⁷m²/s²

We can now calculate the<em> </em><u><em>kinematic</em></u> viscosity of the unknown liquid.

ν=8,32*10⁻⁷m²/s²*900s

ν=7.49*10⁻⁴m²/s

The relationship between the <em>kinematic</em> viscosity and the <em>dynamic</em> viscosity is given by the equation μ=ν * ρ, where μ is the dynamic viscosity and ρ is the density. Thus:

μ=7.49*10⁻⁴m²/s * 970 kg/m³

μ=0.727 kg/m*s

8 0

3 years ago

Read 2 more answers

In this reaction, what roll does the lead (II) nitrate play when 50.0 mL of 0.100M iron (III) chloride are mixed with 50.0 mL of

Aloiza [94]

Answer: The roll of lead (II) nitrate is that it is a limiting reactant of the reaction.

Solution : Given,

Molarity of $FeCl_3$ = 0.1 M

Volume of $FeCl_3$ = 50.0 ml = 0.05 L (1 L = 1000 ml)

Molarity of $Pb(NO_3)_2$ = 0.1 M

Volume of $Pb(NO_3)_2$ = 50.0 ml = 0.05 L

First we have to calculate the moles of $FeCl_3$ and $Pb(NO_3)_2$ .

$\text{ Moles of }FeCl_3=\text{ Molarity}\times \text{ Volume in L}$

$\text{ Moles of }FeCl_3=(0.100M)\times (0.05L)=0.005moles$

$\text{ Moles of }Pb(NO_3)_2=\text{ Molarity}\times \text{ Volume in L}$

$\text{ Moles of }Pb(NO_3)_2=(0.100M)\times (0.05L)=0.005moles$

The balanced chemical reaction is,

$2FeCl_3+3Pb(NO_3)_2\rightarrow 2Fe(NO_3)_3+3PbCl_2$

From the balanced chemical equation, we conclude that

3 moles of lead nitrate react with 2 moles of ferric chloride.

Thus 0.005 moles of lead nitrate react with $=\frac{2}{3}\times 0.005=0.0033$ moles of ferric chloride.

Moles of ferric chloride will be left unreacted = 0.005 - 0.0033 =0.0017 moles

Limiting reagent is the reagent in the reaction which limits the formation of product.

Excess reagent is the reagent in the reaction which is in excess and thus remains unreacted.

Therefore, in the given reaction, lead nitrate is the limiting reagent and ferric chloride is the excess reagent.

7 0

3 years ago

Read 2 more answers

OMG, PLEASE HELP LOL. ONLY ANSWER IF YOU KNOW

Vinil7 [7]

<span>chemical energy

</span><span> hydrogen

hope this helps you(:

have a good day

</span>

6 0

4 years ago