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3 years ago

A student gains 1.0 lb in two weeks. How many grams did he gain?

2 answers:

7 0

1lb= 453.592g but then to change to 2 significant figures you go down two of the numbers so here it will be 4 and 5 that stands for 450g. So he gained 450g. I hope that helped

OlgaM077 [116]3 years ago

3 0

1 lb=453.592 grams
So he gained 453.592 pounds

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An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0

2 years ago

A certain chemical reaction releases 27.4 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of re

Fed [463]

A chemical reaction of exothermic kind releases energy in the form of heat and light.

We are given that reactant releases 27.4kJ/g for every gram of reactant consumed.

We are required to find number of grams of reactant that has been consumed

It is also given that 880J of energy is released

Hence, grams of reactants consumed is 880/27.4gm

Disclaimer:

It has been asked by the student to leave the question as mathematical expression.

For further reference:

brainly.com/question/15000187?referrer=searchResults

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4 0

1 year ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

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Energy is the ability to do work

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