If a dozen apples has a mass of 2.0 kg and 0.20 bushel is 1 dozen apples, How many bushels of apples are in 1.0 Kg of apples

1 answer:

7 0

If 1 dozen apples has a mass of 2.0 kg and 0.20 bushel is 1 dozen apples, how many bushels of apples are in 1.0 kg of apples?

0.1 bushels

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What is the concentration in %m/v of a 0.617 M aqueous solution of methanol (MM = 32.04 g/mol)?

svetoff [14.1K]

Answer:

The correct answer is 1.977 % m/v ≅ 2% m/v

Explanation:

We have:

0.617 M = 0.617 moles methanol/ 1 L solution

We need:

%m/v= grams of methanol/100 mL solution

So, first we convert the moles of methanol to grams by using the MM (32.04 g/mol). Then, we multiply by 0,1 to convert the volume in liters to 100 mL by using the ratio: 100 mL= 0.1 L:

0.617 mol / 1 L x 32.04 g/mol 0.1 L/100 mL= 1.977 g/100 mL= %m/v

6 0

2 years ago

We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find

iragen [17]

Explanation:

a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:

$V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V$

Gold $_{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}$

$1 \mu=10^{-6} \mathrm{m}$

First, find the volume of the sample and then find the area of the sample.

$V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3}$ $=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}$

$V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}$

b. Using the provided information from part $a$ ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :