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Contact [7]
2 years ago
13

Rebekah finds that a solution of sugar in water has a volume of 1. 32 fl. Oz. How should she report this amount using the metric

system? (1 fl. Oz. = 29. 6 mL) 0. 403 m 1. 40 qt 37. 5 g 39. 1 mL.
Chemistry
1 answer:
goldfiish [28.3K]2 years ago
4 0

Rebekah reports the volume of sugar solution using metric system as 39.1 mL. Thus, option D is correct.

The metric system has been the measurement system introduced in 1790s. The system has mediated the units for the measurement of volume, mass, length.

The volume has been the space that has been occupied by matter. The volume can be denoted with SI unit L. However, the another accepted unit for volume has been mL.

Thus, the volume of sugar solution has been reported in mL.

The given sugar solution has volume of 1.32 fl. Oz

The volume in mL can be given as:

\rm 1\;fl\;Oz=29.6\;mL\\1.32\;fl\;Oz=1.32\;\times\;29.6\;mL\\1.32\;fl\;Oz=39.1\;mL

The volume of the sugar solution reported by Rebekah on the metric scale has been 39.1 mL. Thus, option D is correct.

For more information about the metric system, refer to the link:

brainly.com/question/25011390

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What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate
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1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of  Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

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Read 2 more answers
A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
Charra [1.4K]

Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

pH = 0.119

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

5 0
3 years ago
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