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pychu [463]
2 years ago
15

Why is it important to keep the two sides of an equation balanced when solving? what other properties do we use to rewrite expre

ssions and equations?.
Chemistry
1 answer:
kompoz [17]2 years ago
4 0

Answer:

<u>Why is it important to keep the two sides of an equation balanced when solving?</u>

If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. When you solve an equation, you find the value of the variable that makes the equation true.

<u>What other properties do we use to rewrite expressions and equations?</u>

State of matter

Please vote for Brainliest and I hope this helps!

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They create salt when put together
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How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 degrees Celsius to 80 degrees Celsius? (
lapo4ka [179]
The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
7 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
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