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3 years ago

Please write balanced equations for these reactions! WILL GIVE BRAINLIEST! 50 POINTS

Chemistry

1 answer:

Romashka [77]3 years ago

3 0

Answer:

(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)

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The iupac name this compound

MrRa [10]

Answer:

It's HYDROXIDE

Explanation:

You do not call it as hydroxide ion because ion always have + or -

I hope this helps

HAVE A GOOD DAY!

5 0

3 years ago

Read 2 more answers

A typical person has an average heart rate of 75.0 beats/min. Calculate the following a)How many beats does she have in 7.0 year

hram777 [196]

Let us first convert years to minute.

7 years * (365 days / year) * (24 hours / day) * (60 minutes / hour) = 3,679,200 minutes

a. beats = 75.0 beats/min * 3,679,200 minutes

beats = 275,940,000 beats

Since the original given is 7.0 years (2 significant figures), therefore:

beats = 270,000,000 beats

b beats = 75.0 beats/min * 3,679,200 minutes

beats = 275,940,000 beats

Since the original given is 7.000 years (4 significant figures), therefore:

beats = 275,900,000 beats

5 0

3 years ago

Read 2 more answers

How to do an equation

pav-90 [236]

What do you mean by that Oh wait do you mean like 2×2=4 or 2+2=4 something like that maby?? I am still confused?? XD

7 0

3 years ago

The equilibrium: 2 NO2(g) \Longleftrightarrow&iff; N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forwar

Sergeeva-Olga [200]

Answer:

Initial: forward rate > reverse rate
Equilibrium: forward rate = reverse rate

Explanation:

2NO₂(g) → N₂O₄(g) Kc=4.7

The definition of equilibrium is when the forward rate and the reverse rate are equal.

Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.

7 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago