What is the function of the cell membrane

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago

How many are the natural occurring elements?

sergejj [24]

there are 92 naturally occuring elements

6 0

3 years ago

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0

2 years ago

Anything in red is the question

laila [671]

Answer:

Question 19: the three are molecular compounds.

Question 20: CuSO₄.5H₂O

Explanation:

Question 19.

C₂H₄

HF

H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

C₂H₄: two carbon atoms per four hydrogen atoms

HF: one hydrogen atom per one fluorine atom

H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

Question 20. Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

$\dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36$

From which we can solve for x:

$18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5$

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

CuSO₄.5H₂O

4 0

3 years ago

A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

oksian1 [2.3K]

Answer : The $866.66\mu L$ must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of $3.0mCi/mL$ .

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in $\frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L$

Conversion :

$(1ml=1000\mu L)$

Therefore, the $866.66\mu L$ must be administered.

4 0

2 years ago