What is the velocity of a wave with a frequency of 10 Hz and a wavelength of 2 m?

A student is trying to calculate the density of a ball. She already knows the mass, but she needs to determine the volume as wel

murzikaleks [220]

Answer:

V equals four-thirds times pi times r cubed

Explanation:

Volume = a³ , where a is length of each side. Volume = l × w × h , where l is length, w is width and h is height. Volume = 4/3 πr³ , where r is the radius. Volume = πr²h , where r is the radius and h is the height.

7 0

3 years ago

How many molecules of Mg3N2 (magnesium nitride) are formed when excess Mg (magnesium)

dybincka [34]

Explanation:

3Mg(s) + N2(g) = Mg3N2(s)

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

First find the molecular weight using the Periodic Table.

We find that the atomic mass of magnesium is approximately

24.3g, so the molecular weight is just 24.3g\mol

2. Next we need the mole to mole ratio. As there are 3

magnesiums for 1 magnesium nitride (shown by the coefficients), the

mole to mole ratio is 1 mol Mg3N2\3 mol Mg.

3. We need the amount of the substance, in grams. Since you have not

stated it in the question, I'll just do 10g AS AN EXAMPLE. Note that

depending on the amount, the LIMITING REAGENT MAY DIFFER.

4. Finally, we need the molecular weight of Mg3N2, which we can easily

calculate to be around 100.9\mol.

5. Putting this all together, we have 10gMg⋅ (mol Mg\24.3gMg)

(1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)

the units will cancel to leave gMg3N2 (grams of magnesium nitride):

10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)

(100.9g Mg3N2\mol Mg3N2)

Doing the calculation yields approximately 13.84g.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with 10g of nitrogen yields 36.04g

In conclusion, as we produce less amount of Mg3N2 when we assumed that Mg was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have 10g of both. The answer may vary depending on the amount of each substance.

7 0

3 years ago

Read 2 more answers

What is the Golden Rule of Solubility? A. polar dissolves nonpolar. B. nonpolar dissolves polar. C. like dissolves unlike D. lik

DerKrebs [107]

Answer: D. like dissolves like

Explanation:

The solubility of substances is governed by: Like dissolves like, which states that polar compounds are soluble in polar solvents and non polar compounds are soluble in non polar solvents.

Hydrocarbons are non polar in nature due to less difference between the electronegativities of carbon and hydrogen and thus are soluble in non polar solvents only.

Ionic compounds which are formed by elements with high electronegativity difference are polar in nature and thus dissolve in polar solvents.

Example: $NaCl$ in water.

4 0

3 years ago

Select the correct answer.

vaieri [72.5K]

Answer:

b

Explanation:

compounds ..........

6 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$