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2 years ago

Type the correct answer in the box. Use numerals instead of words.

Chemistry

2 answers:

Murrr4er [49]2 years ago

5 0

Answer:

The amount of energy transferred to the coin is 28.5 joules.

Explanation:

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ozzi2 years ago

5 0

Answer:

The amount of energy transferred to the coin is 28.5 joules.

Explanation:

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A gas cylinder is filled with 4.00 moles of oxygen gas at 300.0 k. the piston is compressed to yield a pressure of 400.0 kpa. wh

laila [671]

We can use the ideal gas law equation to find the volume of the gas.
PV = nRT
P - pressure - 400 kPa
V - volume
n - number of moles - 4.00 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 300.0 K
substituting these values in the equation
400 000 Pa x V = 4.00 mol x 8.314 Jmol⁻¹K⁻¹ x 300.0 K
V = 24.9 dm³
Volume is 24.9 dm³

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3 years ago

if an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 mm in front of the len

Natasha2012 [34]

I think the answer is 7mm but I'm not sure.

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3 years ago

Read 2 more answers

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$

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3 years ago

How many grams of carbon are present in 45.0g of CCl4

pentagon [3]

Molar Mass of CCl4= 12+4(35.5)=154 g/mol g Carbon= 45.0g/(154 g/mol)=0.292 mole

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3 years ago

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