/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 You purchase six oranges that we... [FREE SOLUTION] | 91Ó°ÊÓ

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You purchase six oranges that weigh a total of \(2 \mathrm{Ib}_{\mathrm{f}}\) and 13 ounces. After cutting them open and squeezing all the juice your strength allows into a large measuring cup, you weigh the remaining pulp and orange peels. They weigh 1 Ib \(_{\mathrm{f}}\) and 12 ounces and the total volume of the juice is 1.75 cups. What is the specific gravity of orange juice? State any assumptions you make.

Short Answer

Expert verified
The specific gravity of the orange juice is approximately 1.21.

Step by step solution

01

Convert the weight of the oranges to ounces

This problem is working in ounces, not pounds, so to make the calculations easier, convert the weight of the oranges to ounces. 1 pound equals to 16 ounces. So, 2 pounds of oranges is \(2 \times 16 = 32\) ounces. Adding the 13 ounces, the total weight of the oranges is \(32 + 13 = 45\) ounces.
02

Calculate the weight of the orange juice

The weight of the orange juice equals the weight of the oranges minus the weight of the pulp and peels. To find the weight of the juice, first convert the weight of the pulp and peels to ounces. 1 pound of pulp and peels is 16 ounces, so adding the 12 ounces, the total weight of the pulp and peels is \(16 + 12 = 28\) ounces. Now subtract this from the total weight of the oranges, so the weight of the juice is \(45 - 28 = 17\) ounces.
03

Find the specific gravity

Specific gravity is the weight of a substance divided by the weight of an equal volume of water. In this case, we know that 1 cup of water weighs 8 ounces, so 1.75 cups of water weighs \(1.75 \times 8 = 14\) ounces. Now to compute the specific gravity, divide the weight of the orange juice by the weight of the equal volume of water, so the specific gravity is \(17 / 14 \approx 1.21\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Understanding unit conversion is crucial when dealing with a variety of measurements, as it enables clear communication and accurate calculations across different units of measure.

In the context of specific gravity calculations, unit conversion often plays a vital role. For example, when working with weight and volume measurements that are not aligned (like pounds and ounces versus cups), converting them into a common unit allows for a straightforward comparison. In the provided problem, we had to convert pounds into ounces because the weight of the orange peels and the orange juice were initially given in different units.

Unit conversion typically involves a conversion factor, which is a numerical factor used to multiply or divide a quantity when converting from one unit to another. This process might look simple, but being meticulous with conversion factors is necessary to avoid errors. For example, knowing that 1 pound equals to 16 ounces makes it possible to convert 2 pounds of oranges to ounces by multiplying by the conversion factor (2 pounds * 16 ounces/pound = 32 ounces).

These calculations may also involve more complex units such as converting from metric to imperial systems, or from volume to mass, depending on the substance's density. The key to mastering unit conversion is understanding the relationship between various units and practicing the conversions regularly.
Weight Measurement
Weight measurement is a fundamental aspect of science and everyday life, serving as a basis for numerous calculations and assessments. We measure weight to determine the heaviness of an object, and in scientific terms, it’s the gravitational force exerted on an object's mass.

In our specific gravity calculation exercise, weight measurement was a critical step. We measured the total weight of the oranges before and after juicing, as well as the weight of the juice itself. We saw that these weights were expressed in both pounds and ounces. This is where understanding unit conversions becomes useful – by converting these weights into a single unit (ounces in our case), we were able to perform consistent calculations.

To measure weight accurately, a balance or scale is commonly used. These devices can range from simple spring scales to more complex digital scales that provide readings in various units. When measuring for scientific purposes, it's important to use a properly calibrated scale and to record measurements in a standard unit that aligns with the context of the problem we are solving. Also, always take into account the precision of the measuring device, as this can affect the accuracy of your results.
Volume Measurement
Volume measurement represents the amount of space taken up by a substance, be it solid, liquid, or gas. For liquids and gases, volume is often measured in liters, milliliters, gallons, or cups, depending on the context and region.

In the context of the specific gravity calculation for orange juice, we dealt with measuring the liquid's volume after juicing the oranges. The result was conveniently provided in cups, which is a standard unit for volume in the kitchen but not always used in scientific calculations. This illustrates the ubiquitous nature of volume measurement - it appears in both domestic and scientific settings.

When it comes to measuring the volume of liquids, we use measuring cups, graduated cylinders, or pipettes, each suited for different levels of precision. Graduated cylinders and pipettes are particularly common in laboratory settings for their accuracy. In our exercise, 1.75 cups of orange juice were measured, which was crucial for determining its specific gravity. Knowing the weight of the water volume equivalent to the juice volume enabled us to calculate the specific gravity by providing a comparison. To further understand the relationship between weight and volume, it’s key to grasp the concept of density, which directly ties into the principle of specific gravity.

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Most popular questions from this chapter

The reaction \(A \rightarrow B\) is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: \(C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)\) where \(C_{\mathrm{A} 0}\) is the initial concentration of \(\mathrm{A}\) in the reactor and \(k\) is a constant. (a) If \(C_{\mathrm{A}}\) and \(C_{\mathrm{A} 0}\) are in \(\mathrm{Ib}-\) moles \(/ \mathrm{ft}^{3}\) and \(t\) is in minutes, what are the units of \(k ?\) (b) The following data are taken for \(C_{\mathrm{A}}(t):\) $$\begin{array}{cc}\hline t(\min ) & C_{\mathrm{A}}\left(\mathrm{lb}-\mathrm{mole} / \mathrm{ft}^{3}\right) \\\\\hline 0.5 & 1.02 \\\1.0 & 0.84 \\\1.5 & 0.69 \\\2.0 & 0.56 \\\3.0 & 0.38 \\\ 5.0 & 0.17 \\\10.0 & 0.02 \\\\\hline\end{array}$$ Verify the proposed rate law graphically (first determine what plot should yield a straight line), and calculate \(C_{\mathrm{A} 0}\) and \(k\) (c) Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of \(t\) (seconds). Calculate the molarity at \(t=265 \mathrm{s}\).

A mixture is 10.0 mole \(\%\) methyl alcohol, 75.0 mole \(\%\) methyl acetate \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\right),\) and 15.0 mole \(\%\) acetic acid. Calculate the mass fractions of each compound. What is the average molecular weight of the mixture? What would be the mass (kg) of a sample containing 25.0 kmol of methyl acetate?

A thermostat control with dial markings from 0 to 100 is used to regulate the temperature of an oil bath. A calibration plot on logarithmic coordinates of the temperature, \(T(\text { ( } \mathrm{F} \text { ), versus the dial setting, } R\), is a straight line that passes through the points \(\left(R_{1}=20.0, T_{1}=110.0^{\circ} \mathrm{F}\right)\) and \(\left(R_{2}=40.0, T_{2}=250.0^{\circ} \mathrm{F}.\right)\) (a) Derive an equation for \(T\left(^{\circ} \mathrm{F}\right)\) in terms of \(R\). (b) Estimate the thermostat setting needed to obtain a temperature of \(320^{\circ} \mathrm{F}\). (c) Suppose you set the thermostat to the value of \(R\) calculated in Part (b), and the reading of athermocouple mounted in the bath equilibrates at \(295^{\circ} \mathrm{F}\) instead of \(320^{\circ} \mathrm{F}\). Suggest several possible explanations.

Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, \(\left.\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right),\) has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6 .) In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite \(\left(\mathrm{CaCO}_{3}\right)\) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous \(\mathrm{CaSO}_{4}\) solution. The slurry flows from the crystallizer to a filter in which the particles are collected as a filter cake. The filter cake, which is 95.0 wiff solid gypsum and the remainder CaSO_solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO \(_{4}\) as product. A flowchart and relevant process data are given below. Solids content of slurry leaving crystallizer: \(0.35 \mathrm{kg} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) slurry \(\mathrm{CaSO}_{4}\) content of slurry liquid: \(0.209 \mathrm{g} \mathrm{CaSO}_{4} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) Specific gravities: \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), 2.32 ;\) liquid solutions, 1.05 (a) Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer). (b) Takea basis of one liter of solution leaving the crystallizer and calculate the mass (kg) and volume (L) of solid gypsum, the mass of \(\mathrm{CaSO}_{4}\) in the gypsum, and the mass of \(\mathrm{CaSO}_{4}\) in the liquid solution. (c) Calculate the percentage recovery of \(\mathrm{CaSO}_{4}-\) that is, the percentage of the total \(\mathrm{CaSO}_{4}\) (precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous \(\mathrm{CaSO}_{4}\) (d) List five potential negative consequences of discharging \(\mathrm{H}_{2} \mathrm{SO}_{4}\) into the river passing the plant.

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