Question

1. convert 3.44 × 10⁻² kilograms of powdered sugar into tablespoons.

Explanation:

Step1: Recall unit conversions

We know that \(1\) kilogram \( = 1000\) grams, so first convert kilograms to grams. The mass in kilograms is \(3.44\times 10^{- 2}\space kg\). To convert to grams, multiply by \(1000\) (since \(1\space kg=10^{3}\space g\)):
\(3.44\times 10^{-2}\space kg\times10^{3}\space\frac{g}{kg}=3.44\times 10^{1}\space g = 34.4\space g\)

Step2: Recall density of powdered sugar (approximate)

The density of powdered sugar is approximately \(0.85\space g/cm^{3}\), and we also know that \(1\) tablespoon is approximately \(12.5\space cm^{3}\) (or we can use the conversion that \(1\) gram of powdered sugar is approximately \(\frac{1}{0.85}\space cm^{3}\), and then convert \(cm^{3}\) to tablespoons). But an alternative and more direct conversion (using mass - volume - tablespoon conversion) is: We know that the conversion factor for powdered sugar from grams to tablespoons is approximately \(1\space g\) of powdered sugar \(\approx\frac{1}{12.5\times0.85}\) tablespoons? Wait, no, a better way: The volume of \(1\) tablespoon is about \(15\space mL = 15\space cm^{3}\) (some sources say \(12.5\space cm^{3}\) for US tablespoons, let's use \(12.5\space cm^{3}\) for US tablespoon). The density of powdered sugar \( ho=0.85\space g/cm^{3}\). So mass \(m= ho\times V\), so \(V = \frac{m}{ ho}\). Then number of tablespoons \(n=\frac{V}{V_{tablespoon}}\)

First, calculate volume of \(34.4\space g\) of powdered sugar:
\(V=\frac{34.4\space g}{0.85\space g/cm^{3}}\approx40.47\space cm^{3}\)

Then, number of tablespoons \(n=\frac{40.47\space cm^{3}}{12.5\space cm^{3}/tablespoon}\approx3.24\) tablespoons? Wait, maybe I made a mistake. Wait, actually, the conversion from grams to tablespoons for powdered sugar: \(1\) tablespoon of powdered sugar is approximately \(7\space g\) (this is a more practical conversion factor, since different sources give different values, but a commonly used conversion is \(1\) tablespoon of powdered sugar \(\approx7\space g\))

Let's use the conversion factor: \(1\) tablespoon \(\approx7\space g\) of powdered sugar.

So, if we have \(m = 34.4\space g\), the number of tablespoons \(n=\frac{34.4\space g}{7\space g/tablespoon}\approx4.91\) tablespoons? Wait, no, let's go back to the first conversion. Wait, maybe the problem expects us to use the density of powdered sugar as \(1\space g/cm^{3}\) (approximate) for simplicity, or maybe there is a standard conversion.

Wait, another approach: Let's use the fact that \(1\) kilogram of powdered sugar is approximately \(181.82\) tablespoons (since \(1\space kg = 1000\space g\), and if \(1\) tablespoon is \(5.5\space g\) (wait, different sources: some say \(1\) tablespoon of powdered sugar is \(5.5\space g\), some say \(7\space g\)). Let's check with the initial conversion.

Wait, the mass is \(3.44\times 10^{-2}\space kg=0.0344\space kg\)

If \(1\space kg\) of powdered sugar is approximately \(181.82\) tablespoons (using \(1\) tablespoon \( = 5.5\space g\), since \(1000\div5.5\approx181.82\))

Then, \(0.0344\space kg\times181.82\space\frac{tablespoons}{kg}\approx0.0344\times181.82\approx6.25\) tablespoons? Wait, no, if \(1\) tablespoon is \(7\space g\), then \(1000\div7\approx142.86\) tablespoons per kg. Then \(0.0344\space kg\times142.86\space\frac{tablespoons}{kg}\approx0.0344\times142.86\approx4.91\) tablespoons.

But maybe the problem expects us to use the density of powdered sugar as \(1\space g/cm^{3}\) for simplicity (even though it's not accurate, but for the sake of the problem). Let's assume density \( ho = 1\space g/cm^{3}\), and \(1\) tablespoon \(=15\space cm^{3}\) (volume). Then volume \(V=\frac{m}{ ho}=\frac{34.4\space g}{1\space g/cm^{3}} = 34.4\space cm^{3}\)

Number of tablespoons \(n=\frac{34.4\space cm^{3}}{15\space cm^{3}/tablespoon}\approx2.29\) tablespoons. This is confusing because the conversion factors vary.

Wait, maybe the problem has a standard conversion factor. Let's check the correct conversion:

The correct way is:

1. Convert kilograms to grams: \(3.44\times 10^{-2}\space kg=3.44\times 10^{-2}\times10^{3}\space g = 34.4\space g\)

2. The conversion factor for powdered sugar: \(1\) gram of powdered sugar is approximately \(0.12\) tablespoons (this is a rough conversion factor from cooking measurements: \(1\) tablespoon of powdered sugar is about \(8.33\) grams, so \(1\) gram is about \(\frac{1}{8.33}\approx0.12\) tablespoons)

So, number of tablespoons \(=34.4\space g\times0.12\space\frac{tablespoons}{g}\approx4.13\) tablespoons.

But maybe the problem expects us to use the following:

We know that \(1\) kilogram \( = 1000\) grams, and \(1\) tablespoon of powdered sugar is approximately \(5.5\) grams.

So, first, convert \(3.44\times 10^{-2}\space kg\) to grams: \(3.44\times 10^{-2}\times1000 = 34.4\space g\)

Then, number of tablespoons \(=\frac{34.4\space g}{5.5\space g/tablespoon}\approx6.25\) tablespoons (or \(\frac{34.4}{7}\approx4.91\) if we use \(7\space g\) per tablespoon).

But perhaps the intended conversion is:

Density of powdered sugar \( ho = 0.85\space g/cm^{3}\), \(1\) tablespoon \(=12.5\space cm^{3}\)

Volume \(V=\frac{m}{ ho}=\frac{34.4}{0.85}\approx40.47\space cm^{3}\)

Number of tablespoons \(=\frac{40.47}{12.5}\approx3.24\)

Since the problem is likely expecting a rough conversion, and considering that different sources have different conversion factors, but let's use the most common cooking conversion: \(1\) tablespoon of powdered sugar is about \(7\) grams.

So, \(34.4\space g\div7\space g/tablespoon\approx4.91\approx5\) tablespoons (rounded to a reasonable number).

Wait, maybe I made a mistake in the first step. Let's re - do:

Given \(m = 3.44\times 10^{-2}\space kg\)

\(m=3.44\times 10^{-2}\times10^{3}\space g = 34.4\space g\)

If we use the conversion that \(1\) tablespoon of powdered sugar is \(12.5\space g\) (another common value), then number of tablespoons \(=\frac{34.4}{12.5}=2.752\) tablespoons.

This shows that the answer depends on the conversion factor used. But since this is a problem - solving exercise, let's assume the following standard conversion (from mass to volume to tablespoons, using density \( ho = 0.85\space g/cm^{3}\) and \(1\) tablespoon \( = 15\space cm^{3}\)):

\(V=\frac{m}{ ho}=\frac{34.4}{0.85}\approx40.47\space cm^{3}\)

\(n=\frac{40.47}{15}\approx2.7\) tablespoons.

But maybe the problem has a typo or expects a different approach. Alternatively, maybe the question is about unit conversion in terms of mass - volume with a given density (even though it's not provided, but we can use the standard density of powdered sugar as \(0.85\space g/cm^{3}\) and \(1\) tablespoon \( = 12.5\space cm^{3}\))

So, step - by - step with density:

1. Convert mass to volume: \(V=\frac{m}{ ho}\), \(m = 34.4\space g\), \( ho=0.85\space g/cm^{3}\), so \(V=\frac{34.4}{0.85}\approx40.47\space cm^{3}\)

2. Convert volume to tablespoons: \(1\) tablespoon \( = 12.5\space cm^{3}\), so number of tablespoons \(n=\frac{40.47}{12.5}\approx3.24\)

Rounding to a reasonable decimal, we can say approximately \(3.2\) or \(3\) or \(4\) tablespoons. But since the problem is likely expecting a calculation using the conversion \(1\space kg = 1000\space g\) and \(1\) tablespoon \( = 12.5\space g\) (wrong, but for the sake of calculation):

\(3.44\times 10^{-2}\space kg\times1000\space g/kg = 34.4\space g\)

\(34.4\space g\div12.5\space g/tablespoon = 2.752\) tablespoons.

Answer:

Approximately \(\boldsymbol{3 - 5}\) tablespoons (depending on the conversion factor, a more accurate answer using \(1\) tablespoon \( = 8.33\space g\) of powdered sugar: \(\frac{34.4}{8.33}\approx4.13\) tablespoons, so approximately \(4\) tablespoons)