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3 years ago

Tion 1 of 20

Chemistry

2 answers:

GarryVolchara [31]3 years ago

6 0

Answer:

None of these

Explanation:

The questions asks us to find the amount of sodium chloride formed (in grams) when 1.125 X 10^45 molecules of chlorine gas reacts with sodium.

We start by converting the molecules of chlorine gas to moles using Avogadro's number:

1.125 X 10^45 molecules Cl2 X 1 mol Cl2 / 6.022 X 10^23 = 1.87 X 10^21 mol

From here we can set up a BCA table, assuming that we have excess Sodium to react with: We have to assume that Cl2 is the limiting reactant.

2Na + Cl2 ----> 2NaCl

B . excess 1.87 X 10^21 0

C -3.74 X 10^21 -1.87 X 10^21 + 3.74 X 10^21

A X Amount 0 3.74 X 10^21

Now we have 3.74 X 10^21 moles of NaCl that we have to convert to grams:

We can use the molar mass of sodium chloride to accomplish this:

3.74 X 10^21 moles NaCl X 58.44 g NaCl / 1 mol NaCl = 2.18 X 10^23 g NaCl

Answer: None of these

finlep [7]3 years ago

5 0

Answer:

The mass of NaCl produced is 1.2 * 10^23 grams

Explanation:

Step 1: Data given

Number of molecules of Cl2 - gas = 1.25 * 10^45 molecules

Number of Avogadro = 6.022 * 10^23

Step 2: The balanced equation

2 Na + Cl2 → 2 NaCl

Step 3: Calculate moles of Cl2

Moles Cl2 = 1.25 *10^45 / 6.022*10^23

Moles Cl2 = 2.1 *10^21 moles

Step 4: Calculate moles NaCl

For 2.1 *10^21 moles Na we'll have 2.1*10^21 moles NaCl

Step 5: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

Mass NaCl = 2.1*10^21 moles * 58.44 g/mol

Mass NaCl = 1.2 *10^23 grams

The mass of NaCl produced is 1.2 * 10^23 grams

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sertanlavr [38]

Answer:

Volume is the amount of space that a substance occupies. A graduated cylinder is used to measure liquid volume. The formula used to calculate the volume of a solid object is = L x W x H

Explanation:

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2 years ago

(b) What amount (mol) of H₂ can be produced from the given mass of H₂O?

Luden [163]

8 moles of H 2O are produced.

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<h3>How many moles of H2 are required to make one mole of NH3?</h3>

Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.

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An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).

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8 0

1 year ago

Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).

vova2212 [387]

Energy absorbed by Iron block E (iron) = 460.5 J

Energy absorbed by Copper block E (Copper) = 376.8 J

<u>Explanation:</u>

To find the heat absorbed, we can use the formula as,

q = m c ΔT

Here, Mass = m = 10 g = 0.01 kg

ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C

c = specific heat capacity

c for iron = 460.5 J/kg K

c for copper = 376.8 J/kg K

Plugin the values in the above equation, we will get,

q (iron) = 0.01 kg × 460.5 J/kg K × 100 K

= 460.5 J

q (copper) = 0.01 kg × 376.8 J/kg K × 100 K

= 376.8 J

3 0

3 years ago

A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K

Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

6 0

3 years ago

Read 2 more answers

Calculate the kinetic energy of a 8.8 x 10^13 kg comet moving at a speed of 33.0 km/s. Round your answer to 2 significant digits

g100num [7]

Answer:

$KE=4.79\times 10^{22}\ J$

Explanation:

Given that

Mass ,m= 8.8 x 10¹³

Speed ,v= 33 km/s

We know that 1 km= 1000 m

v= 33 x 10³ m/s

As we know that kinetic energy given as

$KE=\dfrac{1}{2}mv^2$

Now by putting the values in the above equation

$KE=\dfrac{1}{2}\times 8.8\times 10^{13}\times 33000^2$

$KE=4.7916\times 10^{22}\ J$

Therefore we can say that the kinetic energy will be

$KE=4.79\times 10^{22}\ J$

6 0

2 years ago