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2 years ago

Can someone tell me how to do this steps by stepHow many grams is there in 3.65 moles of CuSO4

1 answer:

6 0

1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.

2. atomic mass of
Cu = 63.546
S = 32.065
O = 15.9994

3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.

4. 63.546 g + 32.065 g + ( 4 x 15.9994) = 159.609 g

5. this value 159.609 g is the mass in grams of one mol of CuSO4

6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4

7 then you have the multiply the value of one mol by the number of moles that the problem is asking you

8. 159.609 g x 3.65 = 582.571 g

9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"

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In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

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Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

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3 years ago

Suppose you are trying to find the volume of a box based on the following measurements for the length, width, and height of the

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Answer:

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Explanation:

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3 years ago

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3 years ago

A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p

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Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

1 mol = 6.023x10²³ photons

2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J

Finally we calculate the mass equivalence using the equation:

E=m*c²

m=E/c²

m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

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3 years ago

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to

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Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: $\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }$

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

$56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm$

Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.

$P_{2} =\frac{T_{2} P_{1} }{T_{1} }$

$P_{2}=\frac{(337K)(0.55atm)}{418K}$

P₂=0.44 atm

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3 years ago