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3 years ago

When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?

Chemistry

1 answer:

stepladder [879]3 years ago

8 0

Answer:

$m_{precipitated}=24.8g$

Explanation:

Hello,

In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

$m_{precipitated}=108g-83.2g\\\\m_{precipitated}=24.8g$

Best regards.

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What is 10 times as many as 1 hundred is blank hundred or blank thousand ?

Rama09 [41]

10 times as many as 1 hundred is 10 hundred or 1 thousand.

This is the question of simple multiplications related to a place value system.

The given equation in unit form can be rewritten as:

10 × 1 × 100 = (X × 100) or (Y × 1000)

Arranging in two parts, we get,

10 × 1 × 100 = X × 100 ........(1)

10 × 1 × 100 = Y × 1000 ........(2)

Now, calculating the value of X and Y,

From equation (1), we get,

10 × 1 × 100 = X×100

or, 1000 = X × 100

or, X = 10

Hence, 10 times as many as 1 hundred is 10 hundred.

Similarly, from equation (2). we get,

10 × 1 × 100 = Y × 1000

or, 1000 = Y × 1000

or, Y = 1

Hence, 10 times as many as 1 hundred is 1 thousand.

Therefore, 10 times as many as 1 hundred is 10 hundred or 1 thousand.

To learn more about hundreds, thousands, and place values, visit: brainly.com/question/17785584

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8 0

1 year ago

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

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