How many significant figures are in 31400

Identical heat lamps are arranged to shine on two identical containers, one containing water and one methanol (wood alcohol), so

melisa1 [442]

Answer:

The temperature of the methanol will increase more rapidly.

Explanation:

The premise of your question is incorrect. Methanol has an OH group, so there ARE hydrogen bonds among methanol molecules.

However, the specific heat capacity of methanol is 2.53 J°C⁻¹g⁻¹, while that of water is 4.18 J°C⁻¹g⁻¹.

Thus, it takes 65 % more heat energy to raise the temperature of a given mass of water by 1 °C than it does to raise the temperature of an equal mass of methanol by the same amount.

The two samples are receiving heat energy at the same rate, so the methanol will heat up faster than the water.

5 0

4 years ago

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

4 0

4 years ago

What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol)?

NemiM [27]

the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g

The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn

3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2

the formula is n= mass/M so, now substituting values

m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3

so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g

so mass of aluminum oxide obtained = 1.36g

To learn more about Mass:

brainly.com/question/19694949

#SPJ4

3 0

2 years ago

A 7.337 gram sample of chromium is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 9.595 g. Dete

Thepotemich [5.8K]

<h3>Answer:</h3>

Empirical formula is CrO

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of sample of Chromium as 7.337 gram
Mass of the metal oxide formed as 9.595 g

We are required to determine the empirical formula of the metal oxide.

<h3>Step 1 ; Determine the mass of oxygen used </h3>

Mass of oxygen = Mass of the metal oxide - mass of the metal

= 9.595 g - 7.337 g

= 2.258 g

<h3>Step 2: Determine the moles of chromium and oxygen</h3>

Moles of chromium metal

Molar mass of chromium = 51.996 g/mol

Moles of Chromium = 7.337 g ÷ 51.996 g/mol

= 0.141 moles

Moles of oxygen

Molar mass of oxygen = 16.0 g/mol

Moles of Oxygen = 2.258 g ÷ 16.0 g/mol

= 0.141 moles

<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>

Mole ratio of Chromium to Oxygen

Cr : O

0.141 mol : 0.141 mol

1 : 1

Empirical formula is the simplest whole number ratio of elements in a compound.

Thus the empirical formula of the metal oxide is CrO

6 0

3 years ago

Lead metal is produced by heating solid lead (II) sulfide with solid lead (II) sulfate, resulting in liquid lead and sulfur diox

olya-2409 [2.1K]

Answer:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

Explanation:

The balanced reaction of heating solid lead (II) sulfide with the solid lead (II) sulfate to produce liquid lead and sulfur dioxide gas is shown below:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

In the balance reaction above, all the phases are indicated. This reaction is used for the production of lead metal.

4 0

3 years ago