Scientists frequently use all of the following to support the theory that organisms change over time except.

Cora, an electrician, wraps a copper wire with a thick plastic coating. What is she most likely trying to do? keep the electric

Alekssandra [29.7K]

Answer: keep a current from passing out of the wire

Explanation: plastic is an insulator and will not conduct electricity

3 0

3 years ago

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Which elements are metalloids? Check all that apply.

Finger [1]

<h3>Answer:</h3>

Boron
Silicon
Tellurium

<h3>Explanation:</h3>

Elements in the periodic table are either metals, non-metals or metalloids.
Metals are elements that react by losing electron(s) to attain an octet configuration. They occupy groups I, II and III of the periodic table and the transition metals.
Non-metals, on the other hand, are elements that react by gaining electron(s) to attain a stable configuration. Examples include halogens and noble gases among others.
Metalloids are elements in the periodic table that have the properties of both metals and non-metals.
Other examples of metalloids are arsenic, tellurium, germanium and polonium.

7 0

3 years ago

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In a gas grill, 29 lbs propane C3H8 are

dimulka [17.4K]

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given :

Mass of $C_3H_8$ = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of $C_3H_8$ = 44 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$ .

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_3H_8$ react to give 3 moles of $CO_2$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 3=896.7$ moles of $CO_2$

and,

As, 1 mole of $C_3H_8$ react to give 4 moles of $H_2O$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 4=1195.6$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$ .

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2$

$\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs$

and,

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O$

$\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs$

The total mass of products = Mass of $CO_2$ + Mass of $H_2O$

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.

6 0

3 years ago

Does the mass of the sealed jar and its contents change upon the vaporization of the liquid?

Norma-Jean [14]

The answer is No. That is the mass of the sealed jar and its contents does not change upon the vaporization of the liquid, as according to the “law of conservation of mass” , the mass remain conserved when no matter is escape, the mass will remain constant and here also as the jar is sealed, no matter is escaped thus no mass change will be there.

4 0

3 years ago

In CaCo3 + HCl = CaCl2 + H2O +CO3, how many liters of carbon dioxide gas, measured at STP, can be obtained from 45.0 g of calciu

Anon25 [30]

Answer:

B) 10.1 L

Explanation:

Hello,

In this case, for the given chemical reaction which should be corrected as shown below:

$CaCO_3 + HCl \rightarrow CaCl_2 + H_2O +CO_2$

Since 45.0g of calcium carbonate are used, the produced moles of carbon dioxide, via stoichiometry, are found to be:

$n_{CO_2}=45.0g CaCO_3*\frac{1molCaCO_3}{100gCaCO_3}*\frac{1molCO_2}{1molCaCO_3}=0.45molCO_2$

Finally, since STP conditions are referred to a temperature of 273.15K and 1 atm, the volume, by using the ideal gas equation result:

$V=\frac{nRT}{P}=\frac{0.45mol*0.082\frac{atm*L}{mol*K}*273.15K }{1atm} \\ \\V=10.1L$

So the answer is B) 10.1 L.

Best regards.

6 0

3 years ago