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3 years ago

Why is gold usually found in its pure form

Chemistry

1 answer:

attashe74 [19]3 years ago

5 0

gold is usually found in pure form because it is not reacting with other chemicals naturally.

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A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is pr

s344n2d4d5 [400]

A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg

m = 1.43/0.889 = 1.61 molal

7 0

3 years ago

Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802

Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo ( $Q$ ), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto ( $\bar {Q}$ ), en kilojoules por mol, dividido por su masa molar ( $M$ ), en gramos por mol:

$Q = \frac{\bar Q}{M}$ (1)

A continuación, analizamos cada caso:

Metano

$Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }$

$Q = 50.125\,\frac{kJ}{g}$

1 gramo de metano aporta 50.125 kilojoules.

Octano

$Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }$

$Q = 48.246\,\frac{kJ}{mol}$

1 gramo de metano aporta 48.246 kilojoules.

3 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

When exothermic reactions occur, ______________.

marusya05 [52]

Answer:

heat energy is released into the surrounding

3 0

2 years ago

SOMEONE PLEASE HELPPP

xxTIMURxx [149]

A is the correct awnser Beacuse it ether right kne

5 0

3 years ago