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coldgirl [10]
3 years ago
8

Which of the following solutes will increase the boiling point of water the most? magnesium fluoride (MgF2) potassium bromide (K

Br) sucrose (C12H22O11) lithium chloride (LiCl)
Chemistry
2 answers:
Alborosie3 years ago
8 0

Its Magnesium floride (MgF2) and heres the epic reason why" Correct. Magnesium fluoride will dissociate into the greatest number of particles at a ratio of 1:3. Click Next to continue."

Alecsey [184]3 years ago
5 0
I think its potassium bromide
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2. Which of the following is a mixture?
Ganezh [65]

Hey there!

Your answer is D. salt water  

A mixture has more than one substance.

Salt water is made up of multiple different substances, salt (NaCl) and water (H₂O).

Carbon dioxide (CO₂), water vapor (H₂O), and oxygen gas (O₂) are each just one substance.

Hope this helps!

7 0
3 years ago
1. At what temperature dose water freeze
Rus_ich [418]

Answer:

1. water will freeze at a temperature below 32 degrees fahrenheit 0 degree celsius.

2. Ice will melt at a temperature above 32 degrees fahrenheit 0 degrees celsius.

3. water boils at 212 degrees fahrenheit or 100 degrees celsius.

8 0
2 years ago
10. A 2.36-gram sample of NaHCO3 was completely decomposed in an
Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

  • Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

m_r=m_{NaHCO_3}=2.36 g

We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:

m_p=m_{Na_2CO_3}+m_{H_2CO_3}

Apply the law of mass conservation:

m_r=m_p

Substitute the given variables:

m_{NaHCO_3}=m_{Na_2CO_3}+m_{H_2CO_3}

Rearrange for the mass of carbonic acid:

m_{H_2CO_3}=m_{NaHCO_3}-m_{Na_2CO_3}=2.36 g - 1.57 g=0.79 g

8 0
3 years ago
Water covers over 70% of planet Earth and most of the water is salt water. Which
Serga [27]

Answer:

D. 4

Explanation:

6 0
3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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