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Fittoniya [83]
3 years ago
8

Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?

Chemistry
1 answer:
guajiro [1.7K]3 years ago
3 0
<span>In order to covert a unit, you must know certain number of conversions. In here, the conversion is in unit of length. One meter is equal to 1000 meter. So if it is in cubic form, then the answer of one meter cube is also equal to 1000 cube. Then,

32 m</span>³ (1000 mm/1m)³
<span>or
</span>32 m³ (1000³ mm³/1 m³)
= 3.2 x 10¹⁰ mm³
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1. Calculate the heat change associated with cooling a 350.0 g aluminum bar from
Amanda [17]

Answer:

14175 j heat released.

Explanation:

Given data:

Mass of aluminium = 350.0 g

Initial temperature = 70.0°C

Final temperature = 25.0°C

Specific heat capacity of Aluminium = 0.9 j/g.°C

Heat changed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Heat change:

ΔT = Final temperature - initial temperature

ΔT = 25.0°C - 70°C

ΔT = -45°C

Q = m.c. ΔT

Q = 350 g × 0.9 j/g.°C  × -45°C

Q = -14175 j

7 0
2 years ago
2NaOH + H2So4=______+2H2O<br>​
Dmitry [639]

Answer:

2NaOH + H _{2} SO _{4} = >  \:Na _{2}SO _{4}  +2H _{2} O

7 0
3 years ago
At 570. mm Hg and 25°C, a gas sample ne (in mm Hg) at a volume of 1250 mL and a temperature of 175°C and 25°C, a gas sample has
maw [93]

Answer:

final pressure ( P2) = 467.37 mm Hg

Explanation:

ideal gas:

  • PV = nRT

∴ P1 = 570 mm Hg * ( atm / 760 mm Hg ) = 0.75 atm

∴ T1 = 25 ° C = 298 K

∴ V1 = 1.250 L

∴ R = 0.082 atm L / K mol

⇒ n = P1*V1 / R*T1

⇒ n = (( 0.75 ) * ( 1.25 )) / (( 0.082 ) * ( 298 ))

⇒ n = 0.038 mol gas

∴ T2 = 175 °C ( 448 K )

∴ V2 = 2.270 L

⇒ P2 = nRT2 / V2

⇒ P2 = (( 0.038 ) * ( 0.082 ) * ( 448 )) / 2.270

⇒ P2 = 0.615 atm * ( 760 mm Hg / atm ) = 467.37 mm Hg

5 0
3 years ago
Cuantos moles de CO2 se requieren para reaccionar con 2 moles de Ba (OH)2
EastWind [94]

Answer:

hola soy jess, tu respuesta esta aqui

¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2

2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}

1molBa(OH)

2

1molCO

2

= 2 moles CO₂

Explanation:

espero que pueda ayudarte

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3 0
3 years ago
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

Therefore, the mass of PbSO_4 needed are, 1.515 grams.

6 0
2 years ago
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