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3 years ago

Number of grams in 4 mols of cu(no2)2

2 answers:

8 0

Their are 622.23 grams in 4 moles of Cu(NO2)2. Take the molar masses of all the elements and add them together. Then multiply by four because you have 4 moles not 1 mole.

Hope this helps you :)

Drupady [299]3 years ago

6 0

151.55 x 4 you have to add the molar masses for each element together.

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Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o

Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

$Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}$

Now, we are able to solve for m:

$m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} } = 9.70*10^4 g Na=97 Kg Na$ =97 000 g Na

7 0

3 years ago

Read 2 more answers

The number of electrons in n=1 and n=2 shells of aluminum are

harina [27]

Answer:

n=1 holds two electrons and n=2 holds eight electrons.

Explanation:

Hello

In this case, since the atomic number of aluminum is 13, its electron configuration is:

$Al^{13}: 1s^2,2s^2,2p^6,3s^2,3p^1$

In such a way, we can see that the level n=1 is filled with two electrons since the subshell s is able to hold two electrons and the level n=2 is also filled but with eight electrons as s holds two whereas p holds 6. Moreover, n=3 is holding three electrons.

Best regards.

5 0

3 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

25. 00 ml of a buffer solution contains 0. 500 m hclo and 0. 380 m naclo. if 50. 00 ml of water is added to the buffer, what are

geniusboy [140]

The new concentrations of $HClO$ and $NaClO$ are 0.25M and 19M

Calculation of number of moles of each component,

Molarity of $HClO$ = number of moles/volume in lit = 0. 500 M

Number of moles = molarity of $HClO$ × volume in lit = 0. 500 M× 0.025 L

Number of moles of $HClO$ = 0.0125 mole

Molarity of $NaClO$ = number of moles/volume in lit = 0. 38 M

Number of moles = molarity of $NaClO$ × volume in lit = 0. 38 M× 0.025 L

Number of moles of $NaClO$ = 0.95 mole

Calculation of new concentration at volume 50 ml ( 0.05L)

Molarity of $HClO$ = number of moles/volume in lit = 0.0125 mole/0.05L

Molarity of $HClO$ = 0.25M

Molarity of $NaClO$ = number of moles/volume in lit = 0.95mole/0.05L

Molarity of $NaClO$ = 19 M

learn about Molarity

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5 0

2 years ago

Although chemical digestion occurs throughout the digestive system, it mostly occurs in the: Aesophagus and stomach B) liver and

frez [133]

Answer:

The digestive role of the liver is to produce bile and export it to the duodenum. The gallbladder primarily stores, concentrates, and releases bile. The pancreas produces pancreatic juice, which contains digestive enzymes and bicarbonate ions, and delivers it to the duodenum.

Explanation:

The chemical process of digestion begins during chewing as food mixes with ... The esophagus is a tubular organ that connects the mouth to the stomach. ... Acid reflux or “heartburn” occurs when the acidic digestive juices escape into the esophagus. ... The secretions of the liver, pancreas, and gallbladder are regulated by ...

-Both our right just choose one,

6 0

3 years ago