I have two shells joined with a hinge and i use a siphon to filter water to feed myself . To witch group of animals do i belong

Yummy, now if domino’s pizza doesn’t serve this make it your own

Bess [88]

Answer:

thats kinda sus bro

5 0

3 years ago

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Which functions have a vertex with a x-value of 0? select three options.

Murljashka [212]

87 AND (86/) BE BACK

7 0

3 years ago

Kathryn is using cans of juice to fill a cylindrical pitcher that is 11 in. tall and has a radius of 4 in. each can of juice is

Anarel [89]

The number of cans required by Kathryn to fill the pitcher cylinder is 7.3 cans.

<h3>What is volume of cylinder?</h3>

The volume of cylinder is given as the space occupied in the interior of the cylinder. It is given by:

$\rm Volume\;of\;cylinder=\pi r^2h$

The volume of the cylinder pitcher to be filled is:

$\rm Volume\;pitcher=\pi\;\times\;4^2\;\times\;11\;inch^3\\Volume\;pitcher=176\pi inch^3$

The volume of one small can is:

$\rm Volume\;can=\pi\;\times\;2^2\;\times\;6\;inch^3\\Volume\;can=24\;\pi\;inch^3$

The volume given by one can is 24 times pi. The number of cans required to fill the pitcher is:

$\begin{aligned} \rm Number\;of\;cans & =\rm \dfrac{Total\;volume}{Volume\;of\;can}\\\\\rm &=\rm \dfrac{176\;\pi\;inch^3}{24\;\pi\;inch^3}\\\\&=\rm 7.3\end{aligned}$

Kathryn in order to fill the cylinder required 7.3 cans.

Learn more about volume, here:

brainly.com/question/1578538

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4 0

2 years ago

How many grams of lead product would theoretically be produced between 17. 0 g potassium iodide, ki, and 25. 0 g of lead (ii) ni

Fantom [35]

The number of grams of lead product that would theoretically be produced from the given reaction is; <u><em>Mass of lead product = 23.6 g</em></u>

Potassium iodide reacting with Lead(II) Nitrate would yield the balanced equation;

2KI + Pb(NO₃)₂ = 2KNO₃ + PbI₂

2moles : 1 mole = 2 moles : 1 mole

From online tables;

Molar mass of KI = 166 g/mol

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

We are given;

Mass of KI = 17 g

Mass of Pb(NO₃)₂ = 25 g

Thus;

Number of moles of PbI₂ yielded in order to find the limiting reactant will be;

no of moles of PbI₂ from KI = (17g/166 g/mol) × (1 mole of PbI₂/2 mol of KI)

no of moles of PbI₂ from KI = 0.0512 mols

Similarly;

no of moles of PbI₂ from Pb(NO₃)₂ = (25g/331.2 g/mol) × (1 mole of PbI₂/1 mol of Pb(NO₃)₂)

no of moles of PbI₂ from Pb(NO₃)₂ = 0.0755 moles

We can see that KI produced the least amount of moles. Thus;

Mass = 0.0512 * molar mass of PbI₂

From tables, molar mass of PbI₂ = 461 g/mol

Thus;

Mass of lead product = 461 * 0.0512

Mass of lead product = 23.6 g

Read more about theoretical yield at; brainly.com/question/21091465

5 0

2 years ago

I'm bored so tell me a random fact about u or whatever

Tanzania [10]

Answer:

Hai-

Explanation:

I can twist my hand in a 180°-360° movement sooo uhh yea-

6 0

3 years ago

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