Heat is energy, that can be transferred to another item, which is “less hot”.
Answer:
Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).
Step 1
Use dimensional analysis to convert the number of atoms to moles.
1 mole atoms = 6.022 × 10²³ atoms
n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag
Step 2
Convert the moles of Ag to mass.
mass (m) = moles (n) × molar mass (M)
n(Ag) = 3.8193 mol Ag
M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag
m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures
The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.
Explanation:
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

The molar mass will be thus:

Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

The percent composition of Carbon is thus 12.7 %.
<h2><u><em>AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA</em></u></h2>
Answer:
20.25 W
Explanation:
Applying,
P = I²R.................... Equation 1
Where P = Power, I = current drawn by the test light, R = Resistance of the test light
From the question,
Given: I = 1.5 A, R = 9.00 Ω
Substitute these values into equation 1
P = (1.5²)(9)
P = 2.25×9
P = 20.25 W