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2 years ago

3. How would you expect the density of the gummy bear to change if you soaked it in isopropanol (rubbing alcohol, density

Chemistry

1 answer:

77julia77 [94]2 years ago

7 0

Change occurs in the density of the gummy bear to change if we soaked it in isopropanol.

<h3>Effects on density by soaking isopropanol</h3>

We expect the density of the gummy bear to change if we soaked it in isopropanol because the gummy bear absorb the liquid which increase its density due to increasing weight of the gummy bear.

If a substance gains weight then its density also increases and we know that when the gummy bear was soaked in the isopropanol then its weight increases so its density is also changes so we can conclude that change occurs in the density of the gummy bear to change if we soaked it in isopropanol.

Learn more about density here: brainly.com/question/1354972

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The wavelength of transmission is 9.7mm. What is the energy of this transition

Sonja [21]

Answer:

E = 2.05×10⁻²³ J

Explanation:

Given data:

Wavelength of transmission = 9.7 mm (9.7/1000 = 0.0097 m)

Energy of transition = ?

Solution:

9.7 mm (9.7/1000 = 0.0097 m)

9.7×10⁻³ m

Formula:

E = hc/λ

By putting values,

E = 6.63 ×10⁻³⁴ Js × 3×10⁸ m/s / 9.7×10⁻³ m

E = 19.89×10⁻²⁶ J.m / 9.7×10⁻³ m

E = 2.05×10⁻²³ J

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3 years ago

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0

3 years ago

At 300.0 K and 0.987 atm pressure, what will be the volume of 2.30 mol of Ne?

Nadya [2.5K]

Answer:

V = 57.39 L

Explanation:

Given that,

Temperature, T = 300 K

Pressure, P = 0.987 atm

No. of moles of Ne, n = 2.30 mol

We need to find the volume of Ne. We know that, the ideal gas law is as follows :

PV = nRT

Where

P is pressure and R is gas constant

$V=\dfrac{nRT}{P}\\\\V=\dfrac{2.3\times 0.0821\times 300}{0.987 }\\\\V=57.39\ L$

So, the volume of the Ne is 57.39 L.

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3 years ago

What is a colligative property?

V125BC [204]

<span>Colligative properties are properties of solutions that depend on the number of molecules [or ions] in a given volume of solvent and not on the properties (e.g. size or mass) of the compound. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure.</span>

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3 years ago

What are the properties of salt

GrogVix [38]

The compound is sodium chloride

6 0

3 years ago

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