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2 years ago

8

Gold is used for making ornaments give reason

2 answers:

Lena [83]2 years ago

8 0

Answer:

they are nice to look at....

Explanation:

Gold does not tarnish, rust or corrode. Due to its wonderful qualities and its luster, gold is considered the most important metal in jewellery making. As pure gold is too soft for everyday wear, it is alloyed with a mixture of metals in order to make the gold harder, so it can be used for jewellery.

brainliest pls....

Vedmedyk [2.9K]2 years ago

5 0

Answer:

Because looks good for a Christmas tree.

Explanation:

Gold ornaments look pretty shiny and cool to look at.

Hope I helped!

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3 years ago

What are the atoms of Fe(SCN)3

Igoryamba

Answer:

55.845

Explanation:

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2 years ago

Which fire symbol signifies ammunition with a mass explosion?

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The answer is Fire symbol 1

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3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

Calculate the hydrogen-ion concentration (H+] for the aqueous solution in which [OH-1 is 1 x 10-12 mol/L.

Artemon [7]

Answer:If we have [H+][OH-]= Kw = 1.0 x 10^-14

Then [H+]= Kw/ [OH-]= 1.0x 10^-14/ 1 x 10^-11 =1 x 10^-3 mol/L

And here is the solution - as you can see it is an acidic one :

pH = - log [H+]= - log 1 x 10^-3 = 3 < 7

Explanation:

7 0

3 years ago

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