Question

part a 20.690 g cr express your answers using four significant figures n = atoms incorrect; try again; 2 attempts remaining submit previous answers request answer

Explanation:

Step1: Find molar mass of Cr

The molar mass of Chromium (Cr) is approximately \( 51.9961 \, \text{g/mol} \).

Step2: Calculate moles of Cr

Use the formula \( n = \frac{m}{M} \), where \( m = 20.690 \, \text{g} \) and \( M = 51.9961 \, \text{g/mol} \).
\[
n = \frac{20.690}{51.9961} \approx 0.398 \, \text{mol}
\]

Step3: Calculate number of atoms

Use Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \). The formula is \( N = n \times N_A \).
\[
N = 0.398 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 2.397 \times 10^{23} \, \text{atoms}
\]
Rounding to four significant figures, we get \( 2.397 \times 10^{23} \) (or checking the calculation more precisely: \( \frac{20.690}{51.9961}= 0.39827 \); \( 0.39827\times6.022\times 10^{23}= 0.39827\times6.022\times 10^{23}\approx 2.398\times 10^{23} \), but let's do it with more precision:

First, moles: \( \frac{20.690}{51.9961} = 0.3982703 \)

Then atoms: \( 0.3982703\times6.02214076\times 10^{23} \)

\( 0.3982703\times6.02214076 = 0.3982703\times6 + 0.3982703\times0.02214076 \)

\( 2.3896218 + 0.008828 = 2.3984498 \)

So \( 2.398\times 10^{23} \) (four significant figures: 2.398 × 10²³)

Wait, let's check the initial mass: 20.690 g (five significant figures), molar mass of Cr is 51.9961 g/mol (six significant figures). So moles: 20.690 / 51.9961 = 0.398270 (six significant figures). Then multiply by Avogadro's number (6.02214076×10²³):

0.398270 × 6.02214076 = let's compute 0.398270 × 6.02214076:

6.02214076 × 0.4 = 2.408856304, minus 6.02214076 × 0.0017297 = 6.02214076 × 0.001 = 0.00602214, 6.02214076 × 0.0007297 ≈ 0.004405, so total minus ≈ 0.006022 + 0.004405 = 0.010427, so 2.408856 - 0.010427 = 2.398429. So 2.398429 × 10²³, which to four significant figures is 2.398 × 10²³.

Wait, but maybe I made a mistake in the molar mass? Wait, Cr's atomic weight is 51.9961 g/mol (from periodic table). So the calculation is correct.

Wait, let's do it step by step with exact numbers:

Mass (m) = 20.690 g

Molar mass (M) = 51.9961 g/mol

Moles (n) = m / M = 20.690 / 51.9961 ≈ 0.39827 mol

Number of atoms (N) = n × N_A = 0.39827 mol × 6.02214076 × 10²³ atoms/mol

Calculate 0.39827 × 6.02214076:

0.39827 × 6 = 2.38962

0.39827 × 0.02214076 ≈ 0.39827 × 0.02 = 0.0079654, 0.39827 × 0.00214076 ≈ 0.000852, total ≈ 0.0079654 + 0.000852 = 0.0088174

So total N ≈ 2.38962 + 0.0088174 = 2.3984374 × 10²³ atoms

Rounding to four significant figures: 2.398 × 10²³ atoms (since the fourth significant figure is 8, and the next digit is 4, which is less than 5, so we keep it 8? Wait, no: 2.3984374 × 10²³. The first four significant figures are 2, 3, 9, 8. The next digit is 4, so we don't round up. So 2.398 × 10²³.

Alternatively, maybe the question expects using 52.00 g/mol for Cr (four significant figures). Let's try that:

M = 52.00 g/mol

n = 20.690 / 52.00 ≈ 0.39788 mol

N = 0.39788 × 6.022 × 10²³ ≈ 0.39788 × 6.022 = let's calculate:

0.39788 × 6 = 2.38728

0.39788 × 0.022 = 0.00875336

Total ≈ 2.38728 + 0.00875336 = 2.39603336 × 10²³, which is 2.396 × 10²³ (four significant figures). But the molar mass of Cr is 51.9961, which is more precise. So the correct value with four significant figures is approximately 2.398 × 10²³, or maybe 2.397 × 10²³. Wait, let's use a calculator for 20.690 / 51.9961:

20.690 ÷ 51.9961 ≈ 0.39827

0.39827 × 6.02214076 × 10²³ ≈ 0.39827 × 6.02214076 = 2.3984 (as before)

So 2.398 × 10²³ atoms (four significant figures).

Answer:

\( 2.398 \times 10^{23} \) (or if using 52.00 g/mol, 2.396 × 10²³, but the more accurate molar mass gives 2.398 × 10²³)