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3 years ago

What is the formula for copper(ii) phosphate? capitalization and punctuation count!?

1 answer:

4 0

Hello!

The formula for Copper (II) Phosphate is

Cu₃(PO₄)₂

To know how to write this formula we go from left to right. First, Copper (II). It means that the formula has the Cu⁺² ion. Now, Phosphate means that the compound has the PO₄⁻³ ion. Now we combine this two ions, giving to each one the coefficient with the number of the electric charge of the other (e.g. to Cu⁺² we'll give the charge of PO₄⁻³ which is 3 and the result will be Cu₃⁺²).

The combination gives us the result Cu₃(PO₄)₂

Have a nice day!

You might be interested in

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

Consider the following reaction: C6H6 + O2 \longrightarrow ⟶ CO2 + H2O 39.7 grams of C6H6 are allowed to react with 105.7 g of O

Ivenika [448]

Answer:

116.3 grCO2

Explanation:

1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side

C6H6 +15/2 O2⟶ 6CO2 +3 H2O

2nd - we calculate the limiting reagent

39.2gr C6H6*(240grO2/78grC6H6)=120 grO2

we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent

3rd - we use the limiting reagent to calculate the amount of CO2 in grams

105.7grO2*(264grCO2/240grO2)=116.3 grCO2

7 0

3 years ago

What is the balanced form of the following equation?. Br2 + S2O32– + H2O → Br1– + SO42– + H+.

zloy xaker [14]

For the equation to be balanced, the Atom's coefficient on the left side and the right side of the equation has to be equal

so, the answer would be :

Br2 + S2032- + 5H20 -- > BR2- + 2S02- + H+

Hope this helps

8 0

3 years ago

Which process uses acetyl coa as a reactant? apex?

MrRissso [65]

The Krebs cycle uses acetyl CoA as a reactant.

4 0

3 years ago

Dustin is mixing concrete. A formula for concrete calls for 3/8 gal of water. Dustin wants to make 7/9 more concrete than the fo

slava [35]

We can use a ratio to solve this question. Lets refer to the amount formed by the formula as a serving
1 serving needs 3/8 gal
We want to create 7/9 servings extra, so
1 + 7/9 = 16/7
1 : 3/8
16/7 : x
1/(3/8) = (16/7) / x
8 / 3 = 16x / 7
x = 7/6
He needs to use 7/6 gallons of water.

5 0

3 years ago