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Tcecarenko [31]

3 years ago

5

This same pot of water initially started with 100ml and a temperature of 280 C. The water temperature rose to 1030 C. How much e

nergy was absorbed by the water? (Specific heat of water is 1) q=mCΔT

1 answer:

Harlamova29_29 [7]3 years ago

3 0

Answer:The temperature change is 80°C

Explanation:

Heat absorbed by the melting ice (Q) = mL = 312×336 = 104,832 J

∆T = Q/mC = 104,832/312×4.2 = 80°C

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The empirical formula for a compound is C2H4NO. If its molar mass is 232.2 g/mol, what is the molecular formula of the compound?

Irina18 [472]

Empirical formula mass

C2H4NO
2(12)+4(1)+14+16
30+24+4
58g/mol

Molar mass=232.2g/mol

Find n

Molar mass/Empirical formula mass
232.2/58
4

Molecular formula

n×Empirical formula
4(C2H4NO)
C8H16(NO)_4

4 0

2 years ago

People often think bonds store or hold energy. If that is true, should the energy be high or low when a bond forms?

marta [7]

Answer:

The energy should be high.

Explanation:

Bonds do store energy and release it depending if it's endothermic or exothermic. The energy should be low because when a bond forms (endothermic) it releases heat, which helps form bonds. Having a high energy means the bond is absorbing energy, which helps break bonds (endothermic). How this helps!

4 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

Anastasy [175]

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

7 0

3 years ago

Type the correct answer in the box. Spell all words correctly.

cluponka [151]

Answer: hey buddy its not "blome" its biome

The correct term, which is used to describe the water coming from the sky due to the condensation of the water present in the cloud is <u><em>precipitation</em></u>. The precipitation is the water, which comes in form of the rain, hail, snow, drizzle or sleet. This is one of the main sources of the water supply in the biome.

i hope ur looking 4 the one underlined

3 0

3 years ago

PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro

podryga [215]

Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

<u>Answer for Part (2) :</u>

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

3 0

3 years ago