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2 months ago

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Convert 3.4 x 10^23 molecules of NaCl to grams

1 answer:

Alekssandra [3K]2 months ago

7 0

In this case, to find the grams of sodium chloride starting from its molecules, the first step is to determine the moles of sodium chloride by utilizing Avogadro's number. After that, we can obtain grams directly using the molar mass of sodium chloride (58.45 g/mol).

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

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2 months ago

Write the chemical formula for iridium(III) nitride?

lorasvet [2795]

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

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2 months ago

The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um

lions [2927]

$\boxed{\sf 1Å=10^{-10}m}$

$\\ \rm\longmapsto 2.27Å$

$\\ \rm\longmapsto 2.27\times 10^{-10}m$

$\\ \rm\longmapsto 0.227\times 10^{-9}m$

$\\ \rm\longmapsto 0.0227\times 10^{-8}m$

$\\ \rm\longmapsto 0.00227\times 10^{-7}m$

$\\ \rm\longmapsto 0.00023\times 10^{-6}m$

$\\ \rm\longmapsto 0.00023\mu m$

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3 months ago

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

Tems11 [2777]

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

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A sample of a certain binary compound contains 6.93 g of silicon and 7.89 g of oxygen. What is the percent composition of the co

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2 months ago

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