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3 years ago

!!!!! HELP DUE IN 2 MIN !!!

Chemistry

2 answers:

olchik [2.2K]3 years ago

6 0

Answer: 3, 2

Explanation: right on edge 2020

Tpy6a [65]3 years ago

4 0

Answer:

The answer are 3 and 2

Explanation:

just took it

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The pH at the equivalence point will be a. less than 7.00 b. equal to 7.00 Allohalineup of a sto quivalence point of a titration

True [87]

Answer: c. greater than 7.00

Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.

8 0

3 years ago

What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?

MakcuM [25]

The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M

7 0

3 years ago

Read 2 more answers

The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye

ryzh [129]

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

$ln \frac{N}{N_0} = - k*t$ ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

$T _\frac{1}{2} = \frac{ln2}{k}$ ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of $T_\frac{1}{2}$ in above formula (equation 2) :

$28.8 yrs = \frac{ln 2}{ k }$

Multiply both side by k

$28.8 yrs * k = \frac{ln 2 }{k} * k$

Dividing both side by 28.8 yrs

$\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}$

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

$ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t$

$ln (0.0971 ) = -0.0241 yrs ^-^1 * t$

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

$\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}$

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

3 0

3 years ago

Read 2 more answers

The chemical formula of glucose is C6H12O6, so it is classified as a(n)

julsineya [31]

I think it’s classified as a sugar. The ending -ose usually means a sugar.

6 0

3 years ago

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Explain how the processes that occur at mid-ocean ridges and ocean trenches illustrate the cycling of matter.

Nimfa-mama [501]

Explanation:

At one end, new crusts are being produced, at other end the crust is being destroyed and this strikes a unique balance.

At the mid-ocean ridges, the lithospheric plates are diverging. This is implies that the earth is pulling apart here. When the earth is pulling apart, new materials from the asthenosphere comes to the surface thereby creating new lithospheric plate.

As new plates are formed, they push back against the old one. New plates are found very close to the margin and it begins to age away from the margin.

On the other end, old plates are taken away from this center to ocean trenches. At oceanic trenches subduction is occurring.

In a subduction, the lithospheric plate plunges deep into the asthenosphere where they are being melted back.

This is a covergent margin.

This process continues in a dynamic manner to cycle matter on earth.

learn more:

Sea floor spreading brainly.com/question/9912731

#learnwithBrainly

3 0

3 years ago