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3 years ago

10

What is the only object in the table that would sink in corn syrup?

1 answer:

kari74 [83]3 years ago

4 0

Answer:

1.38g/cm 3

Copper-More Dense than 1.38g/cm 3

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Where are the alkali metals and alkaline earth metals and halogens and noble gases located in the periodic table?

mash [69]

The alkali metals are first column, alkali earth 2nd, halogens 2nd to last, and noble gases last. Hope it helps!

7 0

3 years ago

Match: Substance ....... Definition Lewis acid A. Provides H+ in water Bronsted-Lowry base B. Provides OH - in water Arrhenius b

Igoryamba

Answer: F. Electron pair acceptor

Explanation:

A Lewis acid can be properly defined as any substance such as H+ (hygrogen ion) that can accept a pair of electron.

While a Lewis base is any substance such as (OH-) that can donate a pair of electron.

In the neutralization reaction between an acid ( H+ ) and a base (OH-). Hydrogen ion (H+ ) is the Lewis acid because it accepts an electron pair from (OH-).

Other examples of Lewis acid are; Mg2+, K+

3 0

3 years ago

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

$N= N_0(\frac{1}{2})^n$

where

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

here

t = 458,000 years

$t_{\frac{1}{2}}$ = 229,000

$\frac{t}{t_{\frac{1}{2}}}$ = \ $\frac{ 458,000}{229,000}$

n = $\frac{t}{t_{\frac{1}{2}}}$ = 2.000

Now substituting the values

$1.675 = N_0(\frac{1}{2})^{2.000}}$

$1.675 = N_0\times (0.2500)$

$N_0= \frac{1.675}{0.2500}$

$N_0=6.70$

3 0

3 years ago

Why are we not drawing e

alexandr402 [8]

What do you mean by not drawing?

3 0

3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2

nikklg [1K]

Answer : The volume of $NH_4I$ solution required is, 2.93 L

The number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

Explanation :

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

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}$

$\text{Moles of }Pb(NO_3)_2=0.700M\times 0.945L=0.662mol$

Now we have to calculate the moles of $NH_4I$

The balanced chemical reaction is:

$Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react with 2 moles of $NH_4I$

So, 0.662 mole of $Pb(NO_3)_2$ react with $0.662\times 2=1.32$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

$\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}$

$\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L$

Now we have to calculate the moles of $PbI_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react to give 1 moles of $PbI_2$

So, 0.662 mole of $Pb(NO_3)_2$ react to give 0.662 moles of $PbI_2$

Thus, the number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

7 0

3 years ago