What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL

1 answer:

4 0

Density is weight by volume.

First. If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

$\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL$

1L=1000ml

$\frac{1L}{1000mL}$

$(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L$

0.38214 Liters.

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I am confused. Can yall help me answer this question? Also if you do, please show your work.

Eva8 [605]

Answer is: <span>volume of the acid is 0,075 L.
</span>Chemical reaction: KOH + HCl → KCl + H₂O.
V(KOH) = 30 mL · 0,001 L/mL = 0,3 L.
c(KOH) = 0,5 M = 0,5 mol/L.
c(HCl) = 2 M = 2 mol/L.
V(HCl) = ?
From chemical reaction n(KOH) : n(HCl) = 1 : 1.
n(KOH) = n(HCl).
c(KOH) · V(KOH) = c(HCl) · V(HCl).
0,5M · 0,3 L = 2M · V(HCl).
V(HCl) = 0,075 L.

8 0

3 years ago

A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6

Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of $H = 993 \frac{J}{kgC}$

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

$H = \frac{E}{M*(T2-T1)}$