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natima [27]
3 years ago
8

What are the atomic and mass numbers for the periodic table of elements

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
8 0

there you go, hope that helps

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Density = _____<br> (A)weight/length<br> (B)mass/weight<br> (C)mass/volume<br> (D)volume/weight
8_murik_8 [283]
Density= Mass/Volume I am positive I just had an assignment on this
7 0
3 years ago
Read 2 more answers
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

5 0
3 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
A piece of iron is heated to 95.0 celsius and then placed in an insulated vessel containing 250. grams of water at 25.0 celsius.
Nana76 [90]

Answer:

c) 387g

Explanation:

Water;

Mass = 250g

Specific heat = 4.184

Initial Temp, T1 = 25 + 273 = 298K

Final Temp, T2 = 35 + 273 = 308K

Heat = ?

H = mc(T2 - T1)

H = 250 * 4.184 (308 - 298)

H = 10460 J

Iron;

Initial Temp, T2 = 95 + 273 = 368K (Upon converting to kelvin temperature)

Mass = ?

Final Temp, T1 = 35 + 273 = 308

Heat = 10460 (Heat lost by iron is qual to heat gained by water)

Specific heat = 0.45

H = mc(T2-T1)

M = 10460 / [0.45 (308 - 368)]

M = 10460 / 27

M = 387g

7 0
2 years ago
6. What is the oxidation number for the atom indicated in the following compounds.
Hatshy [7]

Answer:

a. +6;

b. +5;

c. +3.

Explanation:

Start with elements with well-known oxidation states.

The oxidation state on oxygen O in compounds is mostly -2. Common exceptions include:

  • -1 in peroxides and
  • positive when oxygen bonds to fluorine.

The oxidation state on group 1 metals (Li, Na, K, etc.) in compounds is mostly +1.

The oxidation state on group 2 metals (Be, Mg, Ca, etc.) in compounds is mostly +2.

Barium Ba is a group 2 metal. The oxidation state on Ba in the compound BaSO₄ is expected to be +2.

The oxidation state on hydrogen H in compounds is mostly +1. The oxidation state on H might be negative when it is bonded to metals.  

The oxidation state on halogens (F, Cl, Br, etc.) is mostly -1. The oxidation state may vary when the halogen is bonded to oxygen or another halogen element.

Compounds are neutral. The oxidation state on all atoms in a compound shall add up to 0. Both BaSO₄ and HClO₂ are neutral.

<h3>BaSO₄</h3>

Oxidation states:

  • Ba: +2;
  • The oxidation state on sulfur S is to be determined;
  • O: -2.

Let the oxidation state on S be x.

2 + x + 4 × (-2) = 0;

x = 6.

Hence, the oxidation state on S in BaSO₄ is +6.

<h3>HClO₂</h3>

Oxidation states:

  • H: +1;
  • Cl here is bonded to oxygen. The oxidation state on chlorine Cl is to be determined;
  • O: -2.

Let the oxidation state on Cl be x.

<em>Refer to the equation in BaSO₄ as an example. Try setting up the equation on your own. </em>

x = 3.

Hence, the oxidation state on Cl is +3.

<h3>PO₄³⁻</h3>

Ions carry charge. Oxidation states on atoms in an ion shall add up to the charge of the ion. The superscript of an ion shows its charge. The superscript 3- in the phosphate ion shows that the ion carries a charge of -3.

Oxidation states:

  • The oxidation state on P is to be found;
  • O: -2.

Let the oxidation state on P be x.

x + 4 × (-2) = -3;

x = 5.

Hence, the oxidation state on P is +5.

4 0
3 years ago
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