Answer:
A piece of gold foil was hit with alpha particles, which have a positive charge. Most alpha particles went right through. This showed that the gold atoms were mostly empty space. Some particles had their paths bent at large angles. A few even bounced backward. The only way this would happen was if the atom had a small, heavy region of positive charge inside it.
c) positively charged and relatively small.
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer: They all transmit energy
Explanation: Because they can all transfer energy, they can transmit it as well. Waves form because particles are disturbed by this transfer of energy.
Answer:
a. MgO(s) + H2CO3(aq) → MgCO3(s) + H2O(l) DOUBLE DISPLACEMENT REACTION.
b. 2KNO3(s)→2KNO2 (s) + O2(8) DESCOMPOSITION REACTION.
c. H2(g) + CuO(s) → Cu(s) + H2O(1) SINGLE DISPLACEMENT REACTION.
d. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(1) COMBUSTION REACTION.
e. H2(8) + Cl2(g) → 2HCl(8) SYNTHESIS REACTION.
f. SO3(g) + H2O(1)→ H2SO4(aq) SYNTHESIS REACTION.