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3 years ago

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100 mL of a strong acid is completely neutralized by 100 mL of a strong base. The observed products are salt and water. Upon fur

ther investigation, the solution has a pH > 7. How is this possible?

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer: The concentration of base is higher than the concentration of the acid.

Explanation:

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The graph shows the number of beans eaten by a random zocco. How many light red kidney beans were eaten on Day 7? How many dark

Tems11 [23]

Answer:

day 7 50 and day 8 40

Explanation:

3 0

3 years ago

Read 2 more answers

Find the number of chromium ions in 3.5 mol of chromium ions.

Ray Of Light [21]

Answer:

2.1 x 10^24 Cr ions

Explanation:

You need to multiply the 3.5 mol by the Avogrado's number (6.022 x 10^23) to get your answer.

3 0

3 years ago

How many moles of H2O are needed to produce 5.6 mol of NaOH?<br> Na2O + H2O --> 2NaOH

navik [9.2K]

Answer: 2.8 moles

Explanation:

The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.

Na2O + H2O --> 2NaOH

1 mole of H2O = 2 moles of NaOH

Let Z moles of H2O = 5.6 mole of NaOH

To get the value of Z, cross multiply

5.6 moles x 1 mole= Z x 2 moles

5.6 = 2Z

Divide both sides by 2

5.6/2 = 2Z/2

2.8 = Z

Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH

3 0

3 years ago

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

3 years ago

If I added 50 grams of salt to water and evaporated the solution with a hot plate, what amount of salt would be left behind?

hjlf

Any salt, even though it reacts with water, will precipitate out completely when the water is completely evaporated. If you start with 50 grams of salt, you will end up with 50 grams of salt.

4 0

4 years ago