When a substance was dissolved in water, the temperature of the water increased. This process is described as

Hard water stains in sinks and showers can be caused by a buildup of solid calcium carbonate. These stains can be removed by bat

LiRa [457]

Answer:

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

Explanation:

The reaction between solid calcium carbonate and a mineral acid such as aqueous HCl is a neutralization reaction and occurs with the evolution of CO2 gas.

The balanced equation is given below

CaCO3 + 2HCl ---> CaCl2 + H20 + CO2

The product CaCl2 is water soluble which accounts for why the stain is removed, while CO2 gas escapes away from the reaction surface.

8 0

3 years ago

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Copper crystallizes in a face-centered cubic lattice (the Cu atoms are at the lattice points and at the face centers). If the de

Delvig [45]

Answer:

The unit cell edge lenght in pm is equal to 361 pm

Explanation:

Data provided:

ρ=Copper density=8.96 g/cm3

Atomic mass of copper=63.54 g/mol

Atoms/cell=4 atoms (in theory)

Avogadro's number=6.02x $10^{23}$ atoms/mol

Since copper has a cubic structure, its cell volume is equal to $a^{3}$ , which can be obtained through the relationship:

cell volume= $\frac{(atoms/cell)(atomic mass)}{(density)(Avogadros number)}$

Substituting the values:

cell volume= $\frac{(4 atoms)(63.54 g/mol)}{(8.96 g/cm3)(6.02x10^{23}) }=4.71x10^{-23}cm^{3}$

clearing, we have:

a= $\sqrt[3]{4.71x10^{-23}cm^{3} }=3.61x10^{-8}cm$

We convert from centimeter to picometer, 1cm=1x $10^{10}$ pm

a= $3.61x10^{-8}cmx\frac{1x10^{10}pm }{1cm} =361 pm$

8 0

3 years ago

How many millimeters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3

KonstantinChe [14]

<span>So we need 0.276 moles of HCl to react. Your concentration is given in moles/liter so 0.276/1.58 = 0.174 liters needed or 174 milliliters</span>

8 0

4 years ago

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

6 0

3 years ago

The modern model of the atom shows that electrons are:

Lina20 [59]

Electrons are orbiting the nucleus in the fxed way paths located in solid sphere

4 0

3 years ago