Help please!!!!!! ASAP

Oduvanchick [21]

The answer is Solar, Wind, Geothermal, Hydrogen, Tidal, Wave, Hydroelectric, Biomass Energy.

7 0

3 years ago

Suppose you were to make a ring from a single strand of gold atoms. How many gold atoms would be required to make such a ring? H

Rudiy27

Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:

54.3 mm = 5.43 cm

Now, we determine the number of gold atoms that will be present in this:

5.43 / 1 x 10⁻⁹

There will be 5.43 x 10⁹ atoms

We now determine the number of moles this is by:

one mole = 6.02 x 10²³ atoms

Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles

The molar mass of gold is 197 g/mol

The mass is 9.01 x 10⁻¹⁵ * 197

The mass of the strand is 1.76 x 10⁻¹² grams

8 0

3 years ago

Read 2 more answers

Calculate the density of the four solutions. All of the solutions has the volume equal to 25 ml. Red solution has 25.0 g of mass

Scilla [17]

Answer:

B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

Explanation:

Density is given as,

D = Mass / Volume

Red Solution,

D = 25 g / 25 mL

D = 1 g/mL

Green Solution,

D = 26.5 g / 25 mL

D = 1.06 g/mL

Yellow Solution,

D = 28.2 g / 25 mL

D = 1.128 g/mL

Blue Solution,

D = 30 g / 25 mL

D = 1.20 g/mL

3 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ΔH° for the r

PolarNik [594]

Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]$

$\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]$

$Delta H=-105kJ$

Therefore, the enthalpy change for this reaction is, -105 kJ

8 0

3 years ago

Explain why the lattice energy of CaSe is approximately 4 times as large as that of KBr. Check all that apply. Check all that ap

Elza [17]

Answer:

According to Coulomb’s law, the Ca and Se ions have 4 times the attractive force (2+ × 2-) than that of the K and Br ions (1+ × 1-).

Explanation:

From Coulomb's law, the attractive force between calcium and selenium ions is four times the attractive force between potassium and bromide ions.

This has something to do with size and magnitude of charge. Calcium ions and selenide ions are smaller and both carry greater charge magnitude than potassium and bromide ions. This paves way for greater electrostatic attraction between them when the distance of the charges apart is minimal. Hence a greater lattice energy.

4 0

3 years ago