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2 years ago

9

An object submerged into 50 cm^3 of water raises the water level to 65 cm^3. The mass of the object is 30 g. What is its density

?

1 answer:

Anna007 [38]2 years ago

8 0

Answer:

Convert the volume to m³.

Explanation:

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A student Increases the temperature of a 417 cubic centimeter balloon from 278 K to 231 K. Assuming constant pressure what shoul

WITCHER [35]

Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K

Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?

Use the general gas equation;

P1*v1/T1 = P2*v2/T2
As, the temperature is constant;
So,
v1/T1 = v2/T2
417/278 = v2/231
v2= 346.5 cm³

5 0

3 years ago

Describe what does happen to the electrons when metal atoms are bonded together.

tankabanditka [31]

Answer: In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions. Metals are shiny.

Explanation: Hope this helped!

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2 years ago

What is photosynthesis?

tamaranim1 [39]

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7 0

3 years ago

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago

In part 2 of the experiment, you will be analyzing a sample of household bleach. A 0.0854 g sample of household bleach is comple

marysya [2.9K]

Answer:

The correct answer is option d.

Explanation:

$NaClO(aq)+2KI(s)+H_2O(l)\rightarrow NaCl(aq)+I_2(aq)+2KOH(aq)$ ..[1]

$I_2(aq)+2Na_2S_2O_3(aq)\rightarrow 2NaI(aq)+Na_2S_4O_6(aq)$ ..[2]

moles of sodium thiosulfate solution = n

Molarity of the sodium thiosulfate solution = 0.150 M

Volume of sodium thiosulfate solution used = 0.890 mL = 0.00089 L

$n=Molarity\times Volume (L)$

$n=0.150 M\times 0.00089 L=0.0001335 mol$

According to reaction [2], 2 moles of sodium thiosulfate reacts with 1 mol $I_2$ .

Then , 0.0001335 mol of sodium thiosulfate will reacts with:

$\frac{1}{2}\times 0.0001335 mol=0.00006675 mol$ of $I_2$

Moles of $I_2$ = 0.00006675 mol

According to reaction [1], 1 mol of $I_2$ is obatnied from 1 mol of NaClO.

Then 0.00006675 mol of $I_2$ will be obtained from:

$\frac{1}{1}\times 0.00006675 mol=0.00006675 mol$

Mass of 0.00006675 mol of NaClO=

0.00006675 mol × 74.44 g/mol = 0.00496887 g

Mass of sample = 0.0854 g

Percentage of NaClO in sample:

$\frac{0.00496887 g}{0.0854 g}\times 100=5.82\%$

4 0

3 years ago