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3 years ago

True or false Bacteria can only be found in a few places on the planet.

Chemistry

1 answer:

frez [133]3 years ago

8 0

Answer:

I think false

Explanation:

It might be false because bacteria can be found on hard/soft surfaces, in our bodies, in the ground, and in the air.

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Write a conversion factor that converts between moles of calcium ions in calcium phosphide, Ca3P2, and moles of Ca3P2.

cricket20 [7]

The conversion factor between moles of calcium ion and moles of $Ca_3P_2$ in $Ca_3P_2$ would be 3 to 1.

<h3>Mole fraction</h3>

$Ca_3P_2$ if formed from 3 carbon atoms and 3 phosphate atoms according to the equation:

$3Ca + 2P --- > Ca_3P_2$

Thus, there are 3 moles of calcium in every 1 mole of $Ca_3P_2$ .

The conversion factor between moles of calcium and moles of $Ca_3P_2$ will. therefore, be 3 to 1 or simply 3:1.

More on mole fractions can be found here: brainly.com/question/8076655

#SPJ1

6 0

2 years ago

Which separation technique could be used to separate a mixture based on differnces in boiling point.

makvit [3.9K]

The answer is B! Chromatography involves solvent separation on a solid medium. Distillation takes advantage of differences in boiling points. Evaporation removes a liquid from a solution to leave a solid material

6 0

3 years ago

Which method preserves an image of a leaf?.

Sav [38]

Answer:

Tree sap flows over the leaf and preserves it.

Explanation:

Amber would preserve the image.

5 0

3 years ago

Read 2 more answers

When solid calcium carbonate is heated, it decomposes according to the reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) Kp = 0.50 A sample o

blagie [28]

Explanation:

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c} [RT]^{\Delta n}$

Given, temperature = $830^{o}C$ = (830 + 273) K = 1103 K

R = 8.314 J/mol K

$\Delta n$ = 1 - 0 = 1

Now, putting the given values into the above formula as follows.

$K_{p} = K_{c} [RT]^{\Delta n}$

0.5 = $K_{c} \times (8.314 \times 1103)^{1}$

$K_{c} = 5.452 \times 10^{-3}$

ICE table for the given reaction will be as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

Initial: c - -

Equilibrium: (c - x) x x

$K_{c}$ = $[CO_{2}]$

$5.452 \times 10^{-3}$ = x

Hence, same amount of CaO is produced.

Moles of CaO = $5.452 \times 10^{-3}$

Mass of CaO = $5.452 \times 10^{-3} \times 57 g/mol$

= 0.310 g

Thus, we can conclude that 0.310 g of CaO produced when equilibrium is established.

8 0

3 years ago

In comparing a 0.25 molality aqueous NaCl solution to a 0.25 molality aqueous CaCl 2 solution:

4vir4ik [10]

Answer:

Explanation:

higher no of particles are in CaCl2 solution

4 0

3 years ago