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2 years ago

PLEASE HELPPP ME

Chemistry

2 answers:

ahrayia [7]2 years ago

7 0

Answer:

1) (5 moles of calcium bromide) x 2 / 1 = 10 moles

2) (2.00 moles of calcium bromide) = 2.00 moles of calcium sulfate

so (2.00) x (136.1 g/mol) = 272.2 g

3) (845 grams of calcium sulfate) / 136.14 g/mol = 6.21 moles

6.21 moles of calcium sulfate = 6.21 moles of calcium bromide

(6.21 moles of calcium bromide) x 199.89 g/mol = 1241.32 g

klasskru [66]2 years ago

6 0

How many miles is in 2NA

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What is a cation and where can you find it on the periodic table?

Maslowich

Answer:

A cation is an ion with fewer electrons than protons. Therefore, it has a positive charge. The electric charge on a proton is equal in amount to the charge on an electron. Anions are atoms or radicals (groups of atoms), that have gained electrons. Since they now have more electrons than protons, anions have a negative charge. Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).

Explanation:

An anion and cation cannot be specifically found on the periodic table.

4 0

2 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

2 years ago

sulfate = SO4 -2 phosphate = PO4 -3 hydroxide = OH- sodium hydroxide aNa(OH)2 bNa(OH)3 cNa2OH 4NaOH

blsea [12.9K]

4. NaOH
____________________________________

6 0

2 years ago

Which sample has more atoms in it? 150.0 g of gold or 10.0 g of lithium.

Deffense [45]

Answer:150g of gold

Explanation: There is a lot more gold and gold is significantly significantly more dense than lithium

6 0

2 years ago

You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (d

Romashka-Z-Leto [24]

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene = 1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

$= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100$

$= (\dfrac{3.8445}{3.8445+1.9229})\times 100$

$= (\dfrac{3.8445}{5.7674})\times 100$

= 66.67%

6 0

2 years ago