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2 years ago

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Describe how a chemist should set up his or her stoichiometric calculation to predict the mass of CuSO4 that forms when a specif

ied mass of CuSO4•5H2O is heated.

2 answers:

stich3 [128]2 years ago

7 0

Convert mass of CuSO4•5H2O to moles using molar mass of CuSO4•5H2O.

Convert moles of CuSO4•5H2O to moles of CuSO4 using a mole ratio.

Convert moles of CuSO4 to mass of CuSO4 using molar mass of CuSO4.

DanielleElmas [232]2 years ago

5 0

1) Calculate the number of moles of Cu SO4 . 5H20 by dividing the specified mass by the molar mass.

2) The ratio of production given by the equation is 1 mol of Cu SO4 . 5 H2O to 1 mol of Cu SO4=> 1:1, meaning that the number of moles of Cu SO4 produced is the same number of moles of Cu SO4.5H20 heated.

3) Finally mutiply the number of moles of Cu SO4 by its molar mass and there you have the mass of Cu SO4 produced.

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Which pair of statements below is correct? Multiple Choice An octet is formed via ionic bonding when one or more valence electro

tresset_1 [31]

Answer:

An octet is formed via ionic bonding when one or more valence electrons are transferred from one atom to another.

An octet is formed via covalent bonding when valence electrons are shared between atoms.

An octet is always formed via ionic bonding

Explanation: The essence of bonding is stability. An octet or duplet state is formed when one or more valence electrons are shared. when the electrons are shared, the type of bond formed is a covalent bond. An octet is formed via ionic bonding when one or more valence electrons are transferred from one atom to another.

3 0

2 years ago

Read 2 more answers

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago

Four different solutions (I, II, III, and IV) are labeled on the pH scale below. A p H scale is shown, fading from red at the le

andreyandreev [35.5K]

Answer: It has the highest number of hydroxide ions

Explanation: I took the review

6 0

2 years ago

Read 2 more answers

The target diol is synthesized in one step from 1-methylcyclopentene, but your lab partner exhausted the supply of that alkene.

andrezito [222]

Answer:

The reagents are $CH_{3}CH_{3}O^{-},OsO_{4},NaHSO_{3}and H_{2}O$ .

Explanation:

1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.

This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.

The chemical reaction is as follows.

4 0

2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = <em>19,26 kJ</em>

I hope it helps!

5 0

2 years ago