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Dima020 [189]
3 years ago
7

a spherical sample has a mass of 2.813g. it has a diameter of 8.0mm. what is the sphere made of (determined by its density as g/

cm *3)
Chemistry
1 answer:
blondinia [14]3 years ago
5 0
To determine the density, we use the following formula:

density= \frac{mass}{volume}

mass= 2.813 grams
volume= ?

we need to first calculate the volume before we can solve for the density. since the question states that the sample has a spherical shape, we can use the volume formula of spheres to find it.

Sphere_{volume}= \frac{4}{3} \pi r^{3}

radius (r)= \frac{diameter}{2}= \frac{8.0 mm}{2} = 4.0 mm

let's plug in the values

V= \frac{4}{3} \pi 4^{3} = 268 mm^3

we need to change the mm^3 to cm^3 using the following conversion

1 cm= 10 mm
268 mm^3 (\frac{1 cm}{10 mm} )^3 = 0.268 cm^3

now we can find the density. 

density= 2.813 grams/ 0.268 cm= 10.5 g/cm3

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dusya [7]

Answer:

True

Explanation:

They are built from particles and blobs of congealed lava ejected from a single vent. As the gas-charged lava is blown violently into the air, it breaks into small fragments that solidify and fall as cinders around the vent to form a circular or oval cone, making them the simplest out there

7 0
3 years ago
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The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

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In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
Energy removal is illustrated in. A. Changing water ice to waterB. Changing water to steamC. Boiling of gasolineD. Evaporation o
Kryger [21]

Answer:

D. Evaporation of sea water

Explanation:

6 0
2 years ago
A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

n = 0.146 moles

for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
3 years ago
Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is __________.a) C2b) N2c) F2
Len [333]

Answer:

B.

Explanation:

ive done this before trust

4 0
2 years ago
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