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3 years ago

I bet half these people are kids from lhs rite now lol

Chemistry

2 answers:

Nana76 [90]3 years ago

7 0

I bet that half of them can type or write better than you can.

Sunny_sXe [5.5K]3 years ago

3 0

No not really this is a global thing millions use this

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The word eclipse most closely means to

Gennadij [26K]

Answer:

OD) Block the view of

Explanation:

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3 years ago

If an aquatic animal relies on photosynthesis to survive, in which ocean zone would it be most successful?

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It is the epipelagic zone.

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3 years ago

Read 2 more answers

How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?

Nina [5.8K]

11.48-gram of $KCl0_3$ are needed to produce 6.75 Liters of $O_2$ gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=298 K

Putting value in the given equation:

$\frac{PV}{RT}=n$

$n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}$

Moles = 0.3588 moles

Now,

$Moles = \frac{mass}{molar \;mass}$

$0.3588 moles = \frac{mass}{32}$

Mass= 11.48 gram

Hence, 11.48-gram of $KCl0_3$ are needed to produce 6.75 Liters of $O_2$ gas measured at 1.3 atm pressure and 298 K

Learn more about the ideal gas here:

brainly.com/question/27691721

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3 0

2 years ago

Copper(II) sulfate is an active ingredient in tree stump remover. In order to help remove

Eva8 [605]

Answer:

the tree red

Explanation:

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3 years ago

How many atoms of Lead are in a 62.7 g sample of Lead(II) permanganate?

Alexandra [31]

Answer:

$\large \boxed {8.48 \times 10^{22}\text{ atoms}}$

Explanation:

1 mol of Pb(MnO₄)₂contains 1 mol of Pb atoms.

1. Moles of Pb

$\text{Moles of Pb} = \text{62.7 g Pb(MnO$_{4}$)}_{2} \times \dfrac{\text{1 mol Pb(MnO$_{4}$)}_{2}}{\text{445.07 g Pb(MnO$_{4}$)}_{2}}\\\\\times \dfrac{\text{1 mol Pb}}{\text{1 mol Pb(MnO$_{4}$)}_{2}} = \text{0.1409 mol Pb}$

2. Atoms of Pb

$\text{Atoms of Pb} = \text{0.1409 mol Pb } \times \dfrac{ 6.022 \times 10^{23}\text{ atoms Pb }}{\text{1 mol Pb }}\\\\= \large \boxed {\mathbf{8.48 \times 10^{22}}\textbf{ atoms Pb }}$

8 0

3 years ago