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3 years ago

13

What is Potassium Oxalate name formula?

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

formula pottasium Oxalate C2k204

Explanation:

C2k204

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How many moles are in 63 grams of lithium nitrite, LiNO2?

marshall27 [118]

Answer:

1.19moles

Explanation:

Li-7

N-14

O2-32

moles*(7+14+32)g/mol=63g

moles=1.19

6 0

2 years ago

Calculate the [ OH − ] and the pH of a solution with an [ H + ] = 0.090 M at 25 °C . [ OH − ] = M pH = Calculate the [ H + ] and

Valentin [98]

Answer: a) $[OH^-]=1.09\times 10^{-13}$ and pH = 1.04

b) $[H^+]=1.02\times 10^{-11}$ and $pH=10.99$

c) $[H^+]=7.08\times 10^{-11}$ and $[OH^-]=1.41\times 10^{-4}$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-\log [OH^-]$

$pH+pOH=14$

a) $[H^+]=0.090M$

$pH=-\log [0.090]=1.04$

$pOH=14-1.04=12.96$

$12.96=-log[OH^-]$

$[OH^-]=1.09\times 10^{-13}$

b) $[OH^-]=0.00098M$

$pOH=-\log [0.00098]=3.01$

$pH=14-3.01=10.99$

$10.99=-log[H^+]$

$[H^+]=1.02\times 10^{-11}$

c) $pH=10.15$

$10.15=-\log [H^+]$

$[H^+]=7.08\times 10^{-11}$

$pOH=14-10.15=3.85$

$3.85=-log[OH^-]$

$[OH^-]=1.41\times 10^{-4}$

3 0

3 years ago

Why do astronauts on the moon seem like they're "Walking on springs" while on earth we are firmly attached to the ground

Sergio [31]

We have gravity here on earth while in space there’s none

8 0

3 years ago

Read 2 more answers

Organic compounds with two oh groups per molecule are called

mafiozo [28]

Glycol
Hope this helps!

4 0

3 years ago

The half-life of plutonium-241 is approximately 13 years. a. How much of a sample weighing 2 g will remain after 70 years? b.

Studentka2010 [4]

Answer: A) 0.0480 grams and B) 56.16 years.

Explanation: Half live is the time in which the amount of radioactive substance remains halve of its initial amount.

The formula we use for solving this type of problem is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount and N is the remaining amount of radioactive substance and n is the number of half lives.

$n=T/t_1_/_2$

where, T is the time and $t_1_/_2$ is half life.

A) from given data, $N_0$ = 2 g

T = 70 years

$t_1_/_2$ = 13 years

N= ?

$n=\frac{70years}{13years}$

n = 5.38

$\frac{N}{2g}=(\frac{1}{2})^5^.^3^8$

$\frac{N}{2g}=0.0240$

N = 0.0480 g

So, 0.0480 grams of the substance will be remaining after 70 years.

B) $N_0$ = 2 g

N = 0.1 g

T = ?

Let's first calculate the value of n for this.

$\frac{0.1}{2}=(\frac{1}{2})^n$

$0.05=0.5^n$

Taking log to both sides:

$log0.05=nlog0.5$

$-1.301=n(-0.3010)$

$n=\frac{1.3010}{0.3010}$

n = 4.32

Half life is 13 years, so we can calculate the time as:

$n=T/t_1_/_2$

$T=n*t_1_/_2$

$T=4.32*13years$

T = 56.16 years

So, it will take 56.16 years for the radioactive substance to decay from 2 g to 0.1 g.

8 0

3 years ago

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