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2 years ago

Calculate the ph of a solution with a [H3O+]=5.6x10-9M

Chemistry

1 answer:

Kisachek [45]2 years ago

6 0

Answer:

pH = 8.25

Explanation:

The acidity or basicity of a solution is measured by its pH. The pH scale ranges from 0 to 14. Solutions having pH from 0-6.9 are considered acidic, at 7 neutral and basic when ranging from 7.1-14.

pH is calculated as,

pH = - log [H⁺] ---- (1)

Where;

[H⁺] = concentration of Acid

Also, for bases pH i calculated using following formula,

pH = 14 - pOH

Therefore, Putting value of H⁺ in equation 1,

pH = - log [5.6 × 10⁻⁹]

pH = 8.25

The solution provided is basic in nature.

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SnO2 + 2 H2 ——> Sn + 2 H2O

SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂ = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

SnO₂ + 2H₂ → Sn + 2H₂O

Ensure that the equation given is balanced;

Now,

the known species is water;

6.02 x 10²³ molecules of water = 1 mole

1.2 x 10²¹ molecules of water = $\frac{1.2 x 10^{21} }{6.02 x 10^{23} }$ = 0.2 x 10⁻²moles

Number of moles of water = 0.002moles

From the balanced chemical equation:

2 mole of water is produced from 1 mole of SnO₂

0.002 moles of water will be produced from $\frac{0.002}{2}$ = 0.001moles

To find the mass;

Mass = number of moles x molar mass

Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol

Mass = 0.001 x 150.7 = 0.15g

3 0

3 years ago

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan

lidiya [134]

Answer:

Second element(Titanium); [Ar] 3d2 4s2

Third element(Vanadium):Ar 3d3 4s2

Explanation:

Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;

Second element(Titanium); [Ar] 3d2 4s2

Third transition element(Vanadium):Ar 3d3 4s2

Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.

6 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

[Rn] 7s25f6d4 Which element is denoted

Colt1911 [192]

Answer:

Explanation:

[Rn] 7s25f6d4

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4 0

3 years ago

What is the mass percentage for copper ?

RideAnS [48]

Answer:

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

3 0

2 years ago