Variable that stays the same

How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas

IrinaVladis [17]

94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the dilution formula

m_1C_1 = m_2C_2

where

m represents the mass and

C represents the percent concentrations

We can rearrange the formula to get

m_2= m_1 × (C_1/C_2)

m_1 = 250 g; C_1 = 35.0 %

m_2 = ?; _____C_2 = 92.5 %

∴ m_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

4 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

3 years ago

How much time would it take for 5.2 x 10^5 atoms of fermium-253 (half-life 3 days) to decay to 6.5 x 104 atoms?

Jlenok [28]

Answer:

9 days, or 3 half-lives

Explanation:

5.2x10^5=520000

6.5x10^4=65000

65000/520000=1/8, or 3 half-lives

3x3=9 days

I'm not the greatest at Chem but this seems more like math than Chem :)

4 0

3 years ago

Use the half-reactions of the reaction Au(OH)3 + HI -> Au +I2 +H2O to answer the questions

kirill115 [55]

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
To enter an electron into a chemical equation use {-} or e
To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

4 0

3 years ago

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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%

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

6 0

3 years ago

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