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3 years ago

A) which substances are the products? ______________________________

1 answer:

5 0

In a chemical reaction the products are found at the right of the equation, the products are what is being made once the reaction is complete. On the right side if the chemical equation is the reactants or starting materials, these are the substances that are combined to provide a product on the right side of the equation. Since I am not able to see the equation, just simple add all the carbons that are on the left and that will tell you how many carbons there are in total on the reactant side and if you add all of the carbons on the right side it will let you know how many carbons there are on the product side. The same steps can be taken for Oxygen.
I am unable to answer the last one as I need more information. But basically the law states that any system for which matter and energy cannot be transfer as it is a closed system, then since the system's mass can't change then it cannot be added or subtracted

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4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.

ikadub [295]

<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
[DA] Multiply/Divide [Cancel out units: $\displaystyle 0.126923 \ mol \ Au$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.126923 mol Au ≈ 0.127 mol Au

3 0

2 years ago

Read 2 more answers

An autoclave is used to sterilize surgical equipment because

avanturin [10]

The answer to your question is D!

5 0

3 years ago

Read 2 more answers

A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the

BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0

3 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Luden [163]

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

To know more about Molar Mass

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8 0

2 years ago

U-235 undergoes many different fission reactions. For one such reaction, when U-235 is struck with a neutron, Ce-144 and Sr-90 a

Lostsunrise [7]

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

The given reaction is $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + x^{1}_{0}n + y^{0}_{-1}\beta$

Now, we balance the mass on both reactant and product side as follows.

235 + 1 = $144 + 90 + (x \times 1) + (y \times 0)$

236 = 234 + x

x = 236 -234

= 2

So, now we balance the charge on both reactant and product side as follows.

92 + 0 = $58 + 38 + (x \times 0) + (y \times -1)$

92 = 96 - y

y = 4

Thus, we can conclude that there are 2 neutrons and 4 beta-particles are produced in the given reaction.

Therefore, reaction equation will be as follows.

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + 2^{1}_{0}n + 4^{0}_{-1}\beta$

8 0

3 years ago