Answer:
y = -2.5 + x
Explanation:
Given the data:
__X____Y
$20 __$17.50
$21 __ $18.50
$22__ $19.50
$23__ $20.50
Using a linear regression calculator :
The linear regression model obtained is :
y = -2.5 + x
Where ; - 2.5 = intercept ; gradient / slope = 1
First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M
Answer:
Solve the following problems (assuming constant temperature). Assume all numbers are 3 sig figs. 1. A sample of oxygen gas occupies a volume of 250 mL at 740 torr pressure. ... the gas exert if the volume was decreased to 2.00 liters? ... A 175 mL sample of neon had its pressure changed from 75.0 kPa to 150 kPa.
Explanation:
Answer:
a. 4,00L
b. 16,00L
c. 12,31L
Explanation:
Avogadro's law says:

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

<em>V₂ = 4,00L</em>
b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

<em>V₂ = 16,00L</em>
c. 25,0g of Ne are:
25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

<em>V₂ = 12,31L</em>
I hope it helps!
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).