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Kaylis [27]
3 years ago
14

Wendy made the following diagram shown to represent the solar eclipse.

Chemistry
2 answers:
DENIUS [597]3 years ago
8 0
Because the moon is not between the sun and earth or i should say earth and the sun.
Pavlova-9 [17]3 years ago
7 0
The moon should be between the sun and Earth
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The flamability (burns) is a physical change ?
NikAS [45]

Flammability is a chemical change because when you burn something, it no longer has the same properties.

5 0
3 years ago
Read 2 more answers
Describe the difference between Al3+<br> and N3-
wariber [46]
Al3+ is cation due to its positve charge
N3- is an anion due to its negative charge
4 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
2 years ago
The mass of a solid substance is 21.112 g. If the volume of the solid substance is 19.5 cm3, calculate the density of the substa
Ostrovityanka [42]

Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

6 0
3 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
azamat
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
                                               ≈ 31.4 mg
5 0
3 years ago
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