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boyakko [2]
3 years ago
5

Waves transfer A density B energy C heat D matter

Chemistry
2 answers:
Airida [17]3 years ago
8 0
It must be energy because there can be different densities, heat, matter in different places
LiRa [457]3 years ago
6 0

Answer:

energy

Explanation:

Waves transfer energy from one place to another

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Which tool is used to measure an object's mass
natka813 [3]

Answer:

A scale, or a weight. Both of which are more scientifically referred to as a balance.

3 0
3 years ago
Read 2 more answers
If i initially have 4.0 l of a gas at a pressure of 1.1 atm, what will the volume be if i increase the pressure of 3.4 atm?
ddd [48]
  The  volume  of  a  gas  that   its  pressure  increase  to  3.4  atm   is    calculated  as   follows

  By  use  of  boyles   law   that  is  P1V1=P2V2
V1=4.0  L
P1=1.1  atm
P2=3.4  atm
V2= P1V1/P2  

(1.1  atm  x  4.0 L)/3.4  atm=  1.29  L
4 0
2 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0
3 years ago
4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
Bas_tet [7]
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0
3 years ago
at 55 years what mass remains of a 200.0 g sample of a radioactive isotope with a half life of 10.0 years
Harman [31]

Answer:

4.419 g

Explanation:

     55 years is 5.5 half lives

200 g   *   (1/2)^5.5 = 4.419 g

3 0
1 year ago
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