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ololo11 [35]
4 years ago
15

Which of the following is soluble in water (A-sand) (B-pepper) (C- oil) (D-honey)​

Chemistry
2 answers:
iren2701 [21]4 years ago
8 0

Answer:

D- Honey

Explanation:

Honey is naturally water-soluble and oil is non polar meaning it is not attracted to water. Sand and pepper are insoluble.

mina [271]4 years ago
6 0

Answer:

The correct answer is (D.)

Explanation:

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I am a gas. I am in the noble gas family and row 1. I have only 2 valence electrons I glow red-orange when in an electric field.
belka [17]
The element is Helium

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3 years ago
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When a 3.00 g 3.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 1.36 kJ ⋅ K − 1
cupoosta [38]

Answer:

Molar heat of solution of KBr is 20.0kJ/mol

Explanation:

Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.

The dissolution of KBr is:

KBr → K⁺ + Br⁻

In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:

q = 1.36kJK⁻¹ × 0.370K

q = 0.5032kJ

Moles of KBr in 3.00g are:

3.00g × (1mol / 119g) = 0.0252moles

Thus, molar heat of solution of KBr is:

0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>

3 0
3 years ago
Describe Write your own caption for this photo.
koban [17]

Answer:

Sit by the fire to warm up

Explanation:

5 0
2 years ago
An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s2? (Formula: F=ma) 0.83 N 1.2 N 11 N 30 N
masya89 [10]
Answer: Option (D) 30N

Detailed Solution:
According to Newton's second law:

F = ma --- (A)

Given:
mass = 5kg
acceleration = 6 m/s^2
F = ?

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F = (5)(6)
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4 0
3 years ago
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You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1
jeka57 [31]

Answer:

\boxed{\text{36 mL}}

Explanation:

1. Write the balanced chemical equation.

\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O

2. Calculate the moles of HCOOH

\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}

3. Calculate the moles of NaOH.

\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}

4. Calculate the volume of NaOH

c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}

3 0
3 years ago
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