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Maru [420]
4 years ago
15

PLEASE ANSWER FAST!!!!!!

Chemistry
2 answers:
Evgen [1.6K]4 years ago
4 0

Answer:   Object A, by 0.5 cubic centimeter

Explanation:

Volume=\frac{Mass}{density}

Hence volume of object A =\frac{\text {Mass of object A}}{\text {density of object A}}=\frac{9g}{10g/cm^3}=0.9cm^3

similarly, volume of object B = \frac{\text {Mass of object B}}{\text {density of object B}}=\frac{4g}{10g/cm^3}=0.4cm^3

Hence from above :  volume of A > volume of B

Difference in volume =(0.9 - 0.4)cm^3=0.5 cm^3

Hence the correct option is Object A has greater volume and  by 0.5 cubic centimeter

lara31 [8.8K]4 years ago
3 0

Answer:

A  by 0,5 cubic centimeters

Explanation:

If you use the density equation:

d=\frac{mass}{volume}→v=\frac{m}{d}

For A:

v=\frac{9[g]}{10g/cm^3} =0,9[cm^3]

For B:

v=\frac{4[g]}{10g/cm^3} =0,4[cm^3]

Doing the difference A-B=0,5 cubic centimeter

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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
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Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

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4 years ago
How many molecular of H2O and O2 are present in 8.5g of H2O2 ?​
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hence H2O is also 0.25 mole i.e.4.5 gm  

O2is 0.125 mole i.e.4 gm

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