<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
D because wants you add the equation it should increase temp but decrease the value in pressure
Answer:
Balance
Measures mass
Metric Ruler
Measures length
Units - centimeter
Spring Scale
Measures weight
Units - grams
Graduated Cylinder
Measures liquid volume
i think this is the right answer please don't report me
1 mm (millimeter) = 0.000001 km (kilometer)
12.5 mm = <span>0.0000125 km
1 mm = </span><span>0.00001 hm (hectometer)
12.5 mm = </span><span>0.000125 hm
1 mm = </span>0.001 m (meter)
12.5 mm = 0.0125 m
1 mm = 0.1 cm (centimeter)
12.5 mm = 1.25 cm
So the only one of the answer choices that doesn't equal 12.5 mm is 0.00125 hm, since 12.5 mm is <span>0.000125 hm.
Answer:
</span><span>0.00125 hm
</span><span>
Hope this helps!</span>
Answer:
596K
Explanation:
Using Charles law equation;
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
V1 = 3.00 L
V2 = double of V1 = 2 × 3.00 = 6.00 L
T1 = 25°C = 25 + 273 = 298K
T2 = ?
Using V1/T1 = V2/T2
3/298 = 6/T2
Cross multiply
298 × 6 = 3 × T2
1788 = 3T2
T2 = 1788 ÷ 3
T2 = 596K
