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4 years ago

C. How many grams of Na2SiO3 can react with 0.800 g of HF?

Chemistry

1 answer:

finlep [7]4 years ago

5 0

Answer:

Mass = 0.610 g

Explanation:

Given data:

Mass of Na₂SiO₃ = ?

Mass of HF = 0.800 g

Solution:

Chemical equation:

Na₂SiO₃ + 8HF → H₂SiF₆ + 2NaF + 3H₂O

Number of moles of HF:

Number of moles = mass/ molar mass

Number of moles = 0.800 g/ 20 g/mol

Number of moles = 0.04 mol

Now we will compare the moles of HF and Na₂SiO₃.

HF : Na₂SiO₃

8 : 1

0.04 : 1//8×0.04 = 0.005

Mass of Na₂SiO₃:

Mass = number of moles × molar mass

Mass = 0.005 mol × 122.07 g/mol

Mass = 0.610 g

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Answer:

False

Explanation:

Temperature is also heat energy, so chemical energy has no affect over it.

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Explain how the cell membrane controls the movement of materials into and out of the cell ?

sukhopar [10]

It's called simple diffusion, the small molecules without charges such as oxygen and carbon dioxide can flow through a plasma membrane without assistance and without expending energy. Other substances such as proteins, glucose and charged particles called ions cannot pass through the selectively permeable membrane.

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3 years ago

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Vikki [24]

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3 years ago

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

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3 years ago