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vredina [299]
3 years ago
12

I am being charged for this service but don't have an account. Need assistance.

Chemistry
2 answers:
Alex787 [66]3 years ago
6 0

Answer:

Email Brainly

Explanation:

Fittoniya [83]3 years ago
4 0

Answer:

go to settings

yvybubuninininininybtvxexexf h j j k injbububyvtc

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using information in titles of appendices b and c, calculate the minimum grams of propane, C3H8 (g), that must be combusted to p
Lubov Fominskaja [6]

<u>Given:</u>

Mass of ice = mass of water = 5.50 kg = 5500 g

Temperature of ice = -20 C

Temperature of water = 75 C

<u>To determine:</u>

Mass of propane required

<u>Explanation:</u>

Heat required to change from ice to water under the specified conditions is:-

q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)

  = m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)

  = 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ

The enthalpy change for the combustion of propane is -2220 kJ/mol

Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane

Molar mass of propane = 44 g/mol

Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g

Ans: 75.15 grams of propane must be combusted.



4 0
4 years ago
Please balance xe+f2=xef6
Marysya12 [62]
If’s chemistry it would be like this- Xe=1 and F=2/Xe=1 and F=6 so, to balance Xe=1 and F=[3x2]/Xe=1 and F=6.
8 0
3 years ago
Pls answer this question and have a AMAGING day!! :))
ladessa [460]

Answer:

B

Explanation:

because it is being cooled down

hoped this helps

3 0
3 years ago
Calculate the change in enthalpy for the reaction. (Use 1.0 g/mL as the density of the solution and 4.18 J/(g⋅∘C) as the specifi
Dovator [93]

Here is the full question

Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH4NO3(s)→NH+4(aq)+NO−3(aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 ∘C and the final temperature (after the solid dissolves) is 21.9 ∘C. Part A Calculate the change in enthalpy for the reaction in kilojoules per mole. (Use 1.0g/mL as the density of the solution and 4.18J/g⋅∘C as the specific heat capacity.) Express your answer to two significant figures and include the appropriate units. ΔHrxn =   ??? kJ/mol

Answer:

26 kJ / mol

Explanation:

Given that;

The mass of NH₄NO₃ = 1.25 g

Number of moles of NH₄NO₃ = Mass of NH₄NO₃ / Molar Mass of NH₄NO₃

Number of moles of NH₄NO₃= 1.25 g / 80.043 g/mol

Number of moles of NH₄NO₃= 0.016 mol

Volume of solution = 25.0 mL

Density of Solution = 1.0g/mL

Since; density = \frac {mass} {volume}

Mass of Solution = Density x Volume

= 1.0g/mL × 25.0mL

= 25 g

Heat Generated (Q)  = mc \delta T

Q= 25g × 4.18 J/g°C x (25.8°C - 21.9°C)

Q =  407.55 J

Q = 407 × 10 ⁻³ kJ

Q = 0.40755 kJ

Δ H_{rxn} = \frac{Heat generated(Q)}{number of molesof NH_4NO_3}

=  \frac {0.40755 kJ}{ 0.016 mol}

= 25.47 kJ/ mol

~ 26 kJ / mol

Therefore, the change in enthalpy for the reaction in kilojoules per mole = 26 kJ / mol

8 0
3 years ago
Particles of a liquid
Tcecarenko [31]

Answer:

A

Explanation:

B. That's a solid

C. Liquid is the only thing that can have viscosity

D. Not necessarily the case

4 0
3 years ago
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