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2 years ago

How do you find the mass of a tissue box? NOT the volume. I need this for an assignment. please help!

Chemistry

1 answer:

Alina [70]2 years ago

8 0

First you need to find the volume
Make sure Density of the box is given .

Now Use density formula

$\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \sf\longmapsto Mass=Density\times Volume$

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How many rings/energy levels would the Bohe model for an element

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It would have THREE rings/energy levels. This is because of the amount of electrons each orbital can hold.

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2 years ago

What are the law of 10 in energy Transfer

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2 years ago

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

$4.67 \times 10^{-6} = x \times (0.10)^{2}$

x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

5 0

2 years ago

Read 2 more answers

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

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2 years ago

Draw the Lewis structure for NF3. What orbitals on N and F overlap to form bonds between these elements

jonny [76]

Is there a picture ????????????

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2 years ago