During a hurricane, what effect can the ocean have on the beach?

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Ivanshal [37]

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

5 0

2 years ago

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of

viktelen [127]

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

Best regards!

4 0

2 years ago

If 21.5 mol of an ideal gas is at 9.65 atm at 10.90 ∘C, what is the volume of the gas?

mamaluj [8]

<h3>Answer:</h3>

51.93 L

<h3>Explanation:</h3>

From the question we are given the following components of an ideal gas;

Number of moles = 21.5 mol

Pressure, P = 9.65 atm

Temperature, T = 10.90°C, but K= °C + 273.15

=284.05 k

We are required to calculate the volume of the ideal gas.

We are going to use the ideal gas equation which is given by;

PV = nRT, where P, V, T and n are the pressure, volume, temperature and moles of the ideal gas respectively. R is the ideal gas constant, 0.082057 L.atm/mol.K

To get the volume, we rearrange the formula to get;

V = nRT ÷ P

= (21.5 × 0.082057 × 284.05 K) ÷ 9.65 atm

= 51.93 L

Thus, the volume of the ideal gas is 51.93 L

3 0

3 years ago

-

ioda

Answer:

02

Explanation:

5 0

3 years ago

PLEASE HELP ASAP!!!

mamaluj [8]

Answer:The approximate atomic mass of lead is 207.24 amu.

Explanation:

Abundance of isotope (I) = 1.4% = 0.014

Atomic mass of the Isotope (I) = number protons + number of neutrons = 82 + 122 = 204 amu

Abundance of isotope (II) = 22.1 % = 0.221

Atomic mass of the Isotope (II) = number protons + number of neutrons = 82 + 125 = 207 amu

Abundance of isotope (III) = 24.1% = 0.241

Atomic mass of the Isotope (III) = number protons + number of neutrons = 82 + 124 = 206 amu

Abundance of isotope (IV) = 52.4% = 0.524

Atomic mass of the Isotope (IV) = number protons + number of neutrons = 82 + 126 = 208 amu

Average atomic mass of an element =

$\sum(\text{atomic mass of an isotopes}\times (\text{fractional abundance)})$

Average atomic mass of lead =

$(204 amu\times 0.014+207 amu\times 0.221 amu+206\times 0.241+208 amu\times 0.524)=207.24 amu$

The approximate atomic mass of lead is 207.24 amu.

7 0

3 years ago