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3 years ago

A form of electricity which can attract things

1 answer:

8 0

A form of electricity which can attract things is static electricity. Static electricity is the result of an imbalance between negative and positive charges in an object. These charges built up o the surface of an object until they find a way to e released or discharged causing the attraction of things.

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How many moles of tin atoms are in a pure tin cup with a mass of 37.6 g ?

Georgia [21]

Tin is an element called Stannum and has the symbol Sn. Molar mass is the mass of 1 mol of a compound, 1 mol of any substance is made of 6.022 x 10²³ units, these units could be atoms making up an element or molecules making up a compound.
While the number of atoms making up 1 mol is the same for any element, the weight of 1 mol of substance varies from one another.
In tin(Sn) molar mass - 118.71 g/mol
In 118.71 g - there's 1 mol of tin
therefore in 37.6 g of tin - 1 x 37.6 / 118.71 = 0.31 mol
In 37.6 g of tin, there's 0.31 mol

3 0

3 years ago

Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden

LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

5 0

3 years ago

At a temperature of 280 K, the gas in a cylinder has a volume of 20.0 liters. If the volume of the gas is decreased to 10.0 lite

seraphim [82]

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

T2 = T1 x V2 / V1

T2 = 280 x 20.0 / 10

8 0

2 years ago

Read 2 more answers

How would your atomic mass prediction have been affected if not all of the water had been evaporated from the calcium carbonate

Yanka [14]

Go to a famlie member, if you can't trust them, trust god. if you cant trust god, BURN IN HELL

6 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago