WHO INVENTED THE LIGHT BOLD, WHO EVER GETS THE ANSWER RIGHT GETS A THANK YOU FROM ME

Chemistry

2 answers:

Romashka [77]3 years ago

8 0

I believe it was thomas edison

aivan3 [116]3 years ago

5 0

This is a hard question because BEING BOLD cant be light ya know... but the light bulb.. thomas edison and joseph swan

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What is the difference between accurate data and reproducible data?

Mkey [24]

Answer:

HERE IS UR ANSWER...

The data which is accurate means it can't be changed further is called accurate data.while,

The data which is changed again and again is called reproducible data.

HOPE IT'S HELP U..

8 0

2 years ago

Identify the formula of the precipitate formed when solutions of sodium perchlorate and nickel(II) nitrate are mixed. If no prec

kumpel [21]

Answer:

2NaClO₄ (aq) + Ni(NO₃)₂ (aq) → 2NaNO₃ (aq) + Ni(ClO₄)₂ (aq)

No reaction

Explanation:

NaClO₄ → Sodium perchlorate

Ni(NO₃)₂ → Nickel (II) nitrate

All the salts from nitrate are soluble salts

All the salts from perchlorate are soluble salts except for the KClO₄

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3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$