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3 years ago

A small 23 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?

Chemistry

1 answer:

Marianna [84]3 years ago

5 0

1/2mv^2
We substitute in the equation above .
1/2•(23)•(1)^2

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How many moles of gas are in 4.4 mL at 1.2 atm?

kirza4 [7]

The Correct Answer Is 3.2

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3 years ago

What is a hot spot? ( Must be in your own words) Pleaseeee Hurryy

Ilia_Sergeevich [38]

Answer:

A hot spot is an area on Earth over a mantle plume or an area under the rocky outer layer of Earth, called the crust, where magma is hotter than surrounding magma. The magma plume causes melting and thinning of the rocky crust and widespread volcanic activity.

Hope this is what you mean be hot spot!

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2 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0

2 years ago

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0

3 years ago

A particular exosolar system has five planets in total: A, B, C, D, and E. The table lists the orbital periods of these planets

sveta [45]

Answer:

The answer is in the picture below

Explanation:

8 0

3 years ago