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love history [14]
3 years ago
5

A material will float on the surface of a liquid if the material has a density less than that of the liquid. Given that the dens

ity of water is approximately 1.0 g/mL, will a block of material having a volume of 1.2 x 10^4 in^3 and weighing 350 lb float or sink when placed in a reservoir of water? [I just don't want the answer, I'd like to know how to solve this, please.]
Chemistry
1 answer:
diamong [38]3 years ago
7 0
Given:
density of water is <span>approximately 1.0 g/mL
block of material
</span><span>volume of 1.2 x 10^4 in^3
and weighing 350 lb
</span>
Required:
Which will float

Solution:
D = M/V where D is density, M is mass and V is volume

convert <span>1.2 x 10^4 in^3 into mL
</span><span>1.2 x 10^4 in^3 (16.39 mL/ 1 in3) = 196,980 mL
</span>
convert <span>350 lb into grams
350 lb (454 grams/1 lb) = 158,900 grams

D = M/V = </span>158,900 grams/ <span>196,980 mL
D = 0.807 g/mL

The block will float in water because it is less dense than water</span>
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Bumek [7]

put 8 in front of the oxygen in the reactants side to make it 16 molecules then put a 5 in front of the co2 in the product side to balance the carbon atoms then put a 6 in front of the H20 on the product side this balances both the hydrogen and oxygen atoms here is a representation

C5H12(g)+8O2(g)=5CO2(g)+6H20

4 0
2 years ago
I need to know the weighted average please!
maria [59]

Answer:

63.5456 amu

Explanation:

0.6917*62.9296+0.3083*64.9278= 63.5456\ amu

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3 years ago
What causes ocean water near the equator to be warmer than ocean water farther north
Elena-2011 [213]
Because the water near the equator gets more direct sunlight.
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List the objects from the smallest to the largest mass.
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2 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
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