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EleoNora [17]
3 years ago
11

A 100-g piece of metal initially at T = 75C is submerged in 100 g of water initially at T = 25C. The specific heat capacity of i

ron is 0.45 J g-1 °C-1 and the specific heat capacity of water is 4.18 J g-1 °C-1. What is the final temperature of both substances in C?
a. 100
b. 75
c. 65
d. 30
e. 25
Chemistry
1 answer:
san4es73 [151]3 years ago
6 0
I’m not so sure but I think it’s b
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The theoretical yield of NaBr from
Dvinal [7]

the percent yield of the reaction is 100%.

The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:

% yield = actual yield / theoretical yield * 100%

% yield of a reaction in this case Rate

In this case, the molar mass of NaBr is 102.9 g / mol, as you know:

444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g

theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g

, Replaced by the definition of percent yield:

percent yield = 728.532 grams / 728.532 grams * 100%

percent yield = 100%

Finally, the percent yield of the reaction is 100%.

<h3 />

FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.

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Is Bromothymol Blue acidic or basic?
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All airplanes have mass; therefore, what force pulls them toward the ground?
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Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

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