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3 years ago

Guys please help me to balance this question❤

Chemistry

1 answer:

zepelin [54]3 years ago

5 0

2 Ca₅(PO₄)₃F + 7 H₂SO₄ → 3 Ca(H₂PO₄)₂ + 7 CaSO₄ + 2 HF

Explanation:

To balance the chemical reaction the number and type of atoms entering the reaction have to be equal to the type and atoms leaving the reaction.

In our case we have:

10 Ca entering the reaction and 10 Ca leaving the reaction

6 PO₄ groups entering the reaction and 6 PO₄ groups leaving the reaction

2 F entering the reaction and 2 F leaving the reaction

7 SO₄ groups entering the reaction and 7 SO₄ groups leaving the reaction

14 H entering the reaction and 14 H leaving the reaction

Learn more about:

balancing chemical equations

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How many oxygen atoms in 4 Al 2 O 3

Zolol [24]

The number of oxygen atoms that are in 4 Al₂O₃ are 12 atoms

Explanation

The subscript in a chemical formula indicate the number of atoms of element immediately before the subscript.

coefficient is the number in front of the formula and tells us how many molecules of a given formula are present.

Since 4Al₂O₃ has coefficient of 4 and the subscript of O is 3 the number of O atoms = 4 x 3=12 atoms

6 0

3 years ago

What is a group in the periodic table of elements?

chubhunter [2.5K]

Answer:

B. The elements that have similar densities and atomic sizes

Explanation:

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3 years ago

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Question 3

harina [27]

Answer: 0.118M

Explanation:

The formula for molarity is: $M=\frac{mol solute}{L solution}$

First, we need to find the number of moles of NaCl.

$2.70gNaCl*\frac{1molNaCl}{58.44gNaCl} =0.05molNaCl$

Next, we must convert millimeters to liters. We can do that by dividing the number of mL by 1000.

$\frac{425mL}{1000} =0.425L$

Now we have our needed data! All we need to do now is plug in our data to the molarity formula.

$M=\frac{0.05mol}{0.425L} =0.118M$

I hope this helps! Pls mark brainliest!! :)

4 0

2 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

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3 years ago

Identify the tropical zones and the temperate zones

dexar [7]

Tropical Zones
*Tropic of Cancer
*Tropic of Capricorn

I cant see it though

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4 years ago

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Guys please help me to balance this question❤​

Guys please help me to balance this question❤