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3 years ago

As a skateboarder moves down a ramp how does her potential energy change

Chemistry

2 answers:

Jlenok [28]3 years ago

7 0

Answer:

Explanation:

yawa3891 [41]3 years ago

4 0

A is correct. Potential energy decreases and turns into kinetic energy.

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5) A1C13 (aq) + Pb(NO3)2 (ag) →

Evgesh-ka [11]

The reaction will produce lead chloride PbCl₂ and aluminium nitrate Al(NO₃)₃.

Explanation:

Aluminium chloride will react with lead nitrate to produce the aluminium nitrate and the lead chloride which is poorly soluble in water.

2 AlCl₃ (aq) + 3 Pb(NO₃)₂ (aq) → 2 Al(NO₃)₃ (aq) + 3 PbCl₂ (s)

where:

(aq) - aqueous

(s) - solid

Learn more about:

similar chemical reactions

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4 0

3 years ago

Molar mass is used in a conversion factor to calculate ___ and uses the units of ____.

inn [45]

Answer:

The molar mass is used in a conversion factor to calculate the weight of one mole of a given compound and uses the units of grams / mole ____.

Explanation:

Se utilizan los pesos atómicos (Z) de cada componente para calcular la masa de 1 mol de compusto, ejemplo:

H20 (agua)

Masa molar H20 = (Peso atomico H) x 2 + Peso atomico 0 = 1, 007g x2 + 16g = 18 gramos / mol

4 0

2 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$