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3 years ago

5

Help!!!

1 answer:

TEA [102]3 years ago

4 0

Is b decrease because it is right

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Need help on this really bad!

timofeeve [1]

Answer:

Help your self

Explanation:

7 0

2 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

An adult human lung has a volume of 2.9 liters. If its volume decreases to 1.2 liters and it contains 0.049 mole of air after it

Lera25 [3.4K]

Answer:

In full volume it contain 0.12 moles.

Explanation:

Given data:

Total volume= Vt = 2.9 L

Decreased volume= Vd = 1.2 L

Number of moles of air present in decreased volume= n = 0.049 mol

Number of moles of air in total volume= n = ?

Solution:

Formula:

Vt/ Vd = n (in total volume) /n ( decreased volume)

2.9 L / 1.2 L = X / 0.049 mol

2.42 = X / 0.049 mol

X = 2.42 × 0.049

X = 0.12 mol

6 0

3 years ago

NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?

Anna007 [38]

Answer: 25.13 g of $N_2O$ and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles$

The balanced chemical equation is:

$NH_4NO_3\rightarrow N_2O+2H_2O$

According to stoichiometry :

1 mole of $NH_4NO_3$ produce = 1 mole of $N_2O$

Thus 0.571 moles of $NH_4NO_3$ will require= $\frac{1}{1}\times 0.571=0.571moles$ of $N_2O$

Mass of $N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g$

1 mole of $NH_4NO_3$ produce = 2 moles of $H_2O$

Thus 0.571 moles of $NH_4NO_3$ will require= $\frac{2}{1}\times 0.571=1.142moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g$

Thus 25.13 g of $N_2O$ and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$

5 0

3 years ago

Determine the mass of oxygen in a 7.2- g sample of A l 2 (S O 4 ) 3 .

astraxan [27]

Molar mass of A l 2 (S O 4 ) 3 = 342.15 g/mol
7.2g/ 342.15 g/mol
you can get number of moles.
then multiply that by12

7 0

3 years ago