All categories

2 years ago

3 why does soldium produce blue colour when dissolve in ammonia?

1 answer:

6 0

Keith may need some help from his manager. A customer is on the phone and she is really angry. She is threatening to post nasty comments on her blog. What should Keith say to his manager? O a) "She's angry. You deal with it." b) "So she's really angry. Her product never arrived. I think we should offer to send her a replacement product free of charge, but wanted your input." "So l'll tell her to calm down. I think she's overreacting. I doubt she'll post c) anything online." "She's really angry; she ordered the product last Wednesday, Don in customer d) service said it would arrive a week later. It was a dollhouse for her niece, Can you take it from here?"

You might be interested in

How many moles are in 400 ml of hydrogen

Nezavi [6.7K]

Answer:

403.176

Explanation:

I hope that the right answer.

8 0

3 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0

2 years ago

Things people think bad about them self

Bumek [7]

1 2 3 4 5 and how the question ask

6 0

3 years ago

How much heat (in kj) is required to warm 13.0 g of ice, initially at -12.0 ∘c, to steam at 113.0 ∘c? the heat capacity of ice i

olga nikolaevna [1]

The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100)
= 39815.88 J
= 39.82 kJ

5 0

3 years ago

Read 2 more answers

What is the percent nitrogen in NaNO3?<br> 56.47%<br> 27.04%<br> 16.48%<br> 14.01%

hichkok12 [17]

I think it’s :answer choice c

3 0

2 years ago

3 why does soldium produce blue colour when dissolve in ammonia?​

3 why does soldium produce blue colour when dissolve in ammonia?