Explanation:
K2O Now the cation is the element at the front so it would be K2 because the 2 belongs with the K which is Potassium and now Anion is the last element O for oxygen
Answer:
CH
Explanation:
We have to obtain the mass of carbon and hydrogen in CO2 and H2O respectively.
For carbon in CO2;
0.845 * 12/44 = 0.23 g
For hydrogen in H20;
0.173 * 2/18 = 0.019 g
We convert the masses to moles of carbon and hydrogen
For carbon - 0.23/ 12 = 0.019 moles
For hydrogen - 0.019/1 = 0.019 moles
Dividing by 0.019 moles for carbon and hydrogen we have the emperical formula of the compound as CH
The mass of I₂ that contains 2.57×10²⁵ molecules is 10843.52 g
From a detailed understanding of Avogadro's hypothesis, we understood 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of I₂ also 6.02×10²³ molecules i.e
<h3>6.02×10²³ molecules = 1 mole of I₂</h3>
Recall:
1 mole of I₂ = 2 × 127 = 254 g
Thus,
<h3>6.02×10²³ molecules = 254 g of I₂</h3>
With the above information, we can obtain the mass of I₂ that contains 2.57×10²⁵ molecules. This is illustrated below:
6.02×10²³ molecules = 254 g of I₂
Therefore,
2.57×10²⁵ molecules = 
<h3>2.57×10²⁵ molecules = 10843.52 g of I₂</h3>
Thus, the mass of I₂ that contains 2.57×10²⁵ molecules is 10843.52 g
Learn more: brainly.com/question/24848191
Answer:
1.07 g of water.
Explanation:
A reaction between an acid and a base makes water and a salt as product.
Our reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Reactants are the acid and the base. Which is the limiting?
2.9 g . 1mol /98 g = 0.0296 moles of acid
3.53 g . 1mol / 40 g = 0.088 moles of base
2 moles of base react to 1 mol of acid
0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid
And we only have 0.0296, sulfuric acid is the limiting
Ratio is 1:2. 1 mol of acid can produce 2 moles of water.
Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.
We convert moles to mass:
0.0592 mol . 18g /mol = 1.07 g
Answer:
a) Xbenzene = 0.283
b) Xtoluene = 0.717
Explanation:
At T = 20°C:
⇒ vapor pressure of benzene (P*b) = 75 torr
⇒ vapor pressure toluene (P*t) = 22 torr
Raoult's law:
∴ Pi: partial pressure of i
∴ Xi: mole fraction
∴ P*i: vapor pressure at T
a) solution: benzene (b) + toluene (t)
∴ Psln = 37 torr; at T=20°C
⇒ Psln = Pb + Pt
∴ Pb = (Xb)*(P*b)
∴ Pt = (Xt)*(P*t)
∴ Xb + Xt = 1
⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)
⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb
⇒ 15 torr = 53 torrXb
⇒ Xb = 15 torr / 53 torr
⇒ Xb = 0.283
b) Xb + Xt = 1
⇒ Xt = 1 - Xb
⇒ Xt = 1 - 0.283
⇒ Xt = 0.717
