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Different "substances" are forms of matter (having volume and taking up space) with different _______________ from each other. a

ExtremeBDS [4]

Answer:

properties

Explanation:

The properties of matter are the various ways in which matter behaves under certain circumstances.

Various substances have different volumes and shapes. These differences account for the various properties of matter.

3 0

3 years ago

AgI + KNO3 What type of reaction is this?

emmasim [6.3K]

Answer:

double replacement

Explanation:

4 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago

This chemical equation represents a chemical reaction.

Strike441 [17]

Explanation:

products are Nacl + H20

5 0

3 years ago

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc

max2010maxim [7]

Answer: The percent yield of the water is 31.98 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

For oxygen gas:

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

4 0

3 years ago