A mixture of gasoline and air explodes when it encounters a spark. this is known as

We see that there are 3 Cl on the products side and 1 on the reactants side, but there are 2 H on the product and 1 on reactant. To fulfill them both, let's put a 6 at HCl.

__Al+ 6HCl→__AlCl₃+ __H₂

Now that we have a 6 at HCl, we can fill in AlCl₃ and H₂.

__Al+ 6HCl→ 2AlCl₃+ 3H₂

All we have left is to fill in Al.

2Al+ 6HCl→ 2AlCl₃+ 3H₂

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__NaCl→ __Na+ __Cl₂

Since we have 2 Cl on the products, we must put 2 on the reactants.

2NaCl→ __Na+ 1Cl₂

With 2 NaCl, we can fill in Na.

2NaCl→ 2Na+ 1Cl₂

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__Na₂S+ __HCl→ __NaCl+ __H₂S

We see 2 Na on reactants, so we can put 2 on the products.

__Na₂S+ __HCl→ 2NaCl+ __H₂S

With 2 H and 2 Cl on the products, we can put a 2 at HCl.

1Na₂S+ 2HCl→ 2NaCl+ 1H₂S

7 0

3 years ago

The density of pure gold is 19.3g/cm3. What is the volume of 1.00 g of pure gold?

soldier1979 [14.2K]

Answer: Volume of 1g of pure gold $v=0.05181 \mathrm{cm}^{3}$

Given;

Density of a pure gold=19.3 $g / c m^{3}$

Mass of a pure gold =1g

To find:

Volume of 1g of pure gold

Solution:

According to the formula,

Density= Mass/Volume

$\rho=m / v$

Where $\rho$ =density of pure gold

m=mass of pure gold

v=volume of pure gold

From the above equation volume can be calculated as

$v=m / \rho$

Substitute the values of mass and density value in the above equation

$v=1 / 19.3$

$v=0.05181 \mathrm{cm}^{3}$

Result:

Thus the volume of 1g of pure gold is $v=0.05181 \mathrm{cm}^{3}$

8 0

3 years ago