___ Is a type of wave that can travel through empty space to transfer energy from place to place

Why is zinc not extracted from ZnO through reduction using CO?

Jet001 [13]

The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO

6 0

1 year ago

A student measures the mass of an 8 cm^3 block of brown sugar to be 12.9 g. What is the density of

Oduvanchick [21]

Answer:

1.6125 g/cm^3

Explanation:

Density= mass/volume

4 0

3 years ago

Atom A and Atom B have the same number of protons and neutrons, but they do not have the same nun ber of

Bogdan [553]

Answer:

The atoms have the same chemical symbol.

6 0

3 years ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago

What happens to the partial pressure of oxygen in a sample of air if the temperature is increased?

ludmilkaskok [199]

Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:

p(O2) = mole fraction of O2 * P(total, air) ----(1)

Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.

6 0

3 years ago

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