What type of acid is sulfuric acid?

a rock weighing 75.0g placed in a graduated cylinder displacing the volume from 31.4 ml to 36.5 ml what is density of the rock i

andrezito [222]

Answer:

ρ≅13.0⋅g⋅mL−1 = 13.0⋅g⋅cm−3

Explanation:

Density=MassPer unit Volume = 75.0⋅g(36.5−31.4)⋅mL =??g⋅mL−1

Note that 1⋅mL = 1⋅cm−3; these are equivalent units of volume;
i.e. 1⋅cm3 = 1×(10−2⋅m)3=1×10−6⋅m3=10−3⋅L=1⋅mL.

7 0

3 years ago

21. The formation of an ionic bond involves the a transfer of electrons. b. transfer of neutrons. c. transfer of protons. d. sha

iren2701 [21]

Answer:

I think its C but if its not try A then

Explanation:

7 0

2 years ago

In which reaction is a precipitation occurring

iogann1982 [59]

It is reacting as a chemical reaction .

6 0

3 years ago

For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of t

makvit [3.9K]

Answer:

$\frac{N_{v}}{N}=1.92*10^{-4}$

Explanation:

First of all we need to find the amount of atoms per volume (m³). We can do this using the density and the molar mass.

$7.40 \frac{g}{cm^{3}}*\frac{1mol}{85.5 g}*\frac{6.023*10^{23}atoms}{1mol}*\frac{1000000 cm^{3}}{1m^{3}}=5.21*10^{28}\frac{atoms}{m^{3}}$

Now, the fraction of vacancies is equal to the N(v)/N ratio.

N(v) is the number of vacancies $1*10^{25}m^{-3}$
N is the number of atoms per volume calculated above.

Therefore:

The fraction of vacancies at 600 °C will be:

$\frac{N_{v}}{N}=\frac{1*10^{25}}{5.21*10^{28}}$

$\frac{N_{v}}{N}=1.92*10^{-4}$

I hope it helps you!

7 0

3 years ago

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular

mars1129 [50]

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

$\frac{actual yield}{theoretical yield} .100$

Where the actual yield is how much product was obtained, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

$\frac{actual yield}{theoretical yield} .100$

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

5 0

2 years ago

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