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3 years ago

9

How do I balance this equation?

1 answer:

gavmur [86]3 years ago

7 0

Answer: place a 2 in front of NaNo3 on left side of equation while leaving the other blanks empty or you can place a 1 in those blanks

Explanation:

Step 1 count and write down the amount of each given element for both sides

Step 2 begin placing numbers (coefficients) to each side to balance

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¿Como pasar 254 meses a minutos?<br> Por favor ayudaaa!! <br> Le doy corona a la respuesta

Dimas [21]

Answer:

Ver las respuestas abajo.

Explanation:

Este problema se puede resolver conociendo la relacion entre horas y minutos, sabemos que:

1 hora [h] → 60 minutos [min]

De esta manera:

2 [min] = 2/60 = 0.033 [h]

15 [min] = 15/60 = 0.25 [h]

30 [min] = 30/60 = 0.5 [h]

10 [min] = 10/60 = 0.166 [h]

6 [min] = 6/60 = 0.1 [h]

20 [min] = 20/60 = 0.33 [h]

5 [min] = 5/60 = 0.0833 [h]

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3 years ago

Important functional groups in biomolecules include --------

Andrew [12]

B nucleic acids I think

4 0

3 years ago

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Zoom in if need to<br> PLEASE HELP DUE TOMARROW

padilas [110]

The best way to do this is to google search each question, or look it up on quizlet.com

5 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

What is the pH of a solution prepared by dissolving<br>0.8 g NaOH in water to make 200 mL solution?

Ad libitum [116K]

Answer:

pH>7

Explanation:

bases tend to increase the pH of a solution. since water has the pH of 7 and NaOH has pHof 14, the overall pH of solution will increase.

hope it's helpful.

3 0

3 years ago

Read 2 more answers