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3 years ago

PLEASE HELP ME! Determine the number of atoms in 4.27 g gold.

1 answer:

6 0

This can be done with some simple dimensional analysis. First you convert your gram of gold into moles of gold There are 196.97 grams of gold for every one mole of gold so you divide 4.27 by 196.67 to get 0.0217 moles of gold. You then multiply the moles of gold by Avogadro's number, 6.02 X 10^23. Avogadro's number indicates the number of atoms in one mole. Therefore, there are 1.307 X 10^22 atoms in 4.27 grams of gold.

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Why do California and Peru both have large deposits of gold?

irinina [24]

Both California and Peru have or have had significant gold deposits because of their proximity to where two tectonic plate boundaries meet. Off the California coast there is the Pacific Plate and the North American Plate which meet at the San Andreas Fault but there also is the Farallon Plate which apparently has subducted somewhat below the North American Plate. Off Peru, the Nazca Plate is subducting under the South American Plate and which has caused the Andean orogeny.

3 0

3 years ago

Do elements of the halogen family lose one electron to become halide ions

lana66690 [7]

Its on Google..."The group number of a representative element in the periodic table is related to the number of valence electrons it has. ... Elements of the halogen family lose one electron to become halide ions."

5 0

3 years ago

There are over 100 elements, but only

elena-14-01-66 [18.8K]

Answer:

Symbol_Element

He_Helium

Li_Lithium

Be_Beryllium

B_Boron

C_Carbon

N_Nitrogen

O_Oxygen

F_Fluorine

Ne_Neon

Na_Sodium

Mg_Magnesium

Al_Aluminum

Si_Silicon

P_Phosphorous

S_Sulfur

Cl_Chlorine

Ar_Argon

K_Potassium

Ca_Calcium

Sc_Scandium

Ti_Titanium

V_Vanadium

Cr_Chromium

Mn_Manganese

Fe_Iron

Co_Cobalt

Ni_Nickel

Cu_Copper

Zn_Zinc

Ga_Gallium

Ge_Germanium

As_Arsenic

Se_Selenium

Br_Bromine

6 0

3 years ago

How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

gulaghasi [49]

Answer:

$\boxed {\boxed {\sf 10.2 \ kJ}}$

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

$q=mc \Delta T$

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

m= 225 g
c= 0.897 J/g° C
ΔT= 50.5 °C

$q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

Multiply the first two numbers. The units of grams cancel.

$q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

$q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)$

$q= (201.825\ J / \textdegree C)(50.5 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 201.825 \ J * 50.5$

$q= 10192.1625 \ J$

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

$\frac { 1 \ kJ}{ 1000 \ J}$

Multiply by the answer we found in Joules.

$10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}$

$10192.1625 * \frac{ 1 \ kJ}{ 1000 }$

$\frac {10192. 1625}{1000} \ kJ$

$10.1921625 \ kJ$

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

$10.2 \ kJ$

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0

2 years ago

A certain chemical reaction releases 36.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of re

Lilit [14]

Answer:

0.038 g of reactant

Explanation:

Data given:

Heat release for each gram of reactant consumption = 36.2 kJ/g

mass of reactant that release 1360 J of heat = ?

Solution:

As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J

As we know

1 KJ = 1000 J

36.2 kJ = 36.2 x 1000 = 36200 J

So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release

Apply unity formula

36200 J of heat release ≅ 1 gram of reactant

1360 J of heat release ≅ X gram of reactant

Do cross multiplication

X gram of reactant = 1 g x 1360 J / 36200 J

X gram of reactant = 0.038 g

So 0.038 g of reactant will produce 1360 J of heat.

5 0

2 years ago