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jenyasd209 [6]
3 years ago
8

What is the answer to question 7, NH4C2H3O2

Chemistry
2 answers:
Veronika [31]3 years ago
7 0

Answer:

Ammonium acetate was not sure at first but it was Ammonium acetate sorry i was confused at first

Hope this helps :D

Sliva [168]3 years ago
4 0
The answer is Ammonium acetate
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Warm water rising to the surface of the
White raven [17]

Answer:

convection

Explanation:

7 0
3 years ago
How is stoichiometry used to calculate the amount of product from amount of reactant?
IceJOKER [234]
<span>Stoichiometry deals with the quantitative measurement of reactants and products in a chemical reaction. Let suppose you are given with following reaction;

                                            A  +  2 B   </span>→    3 C

According to this reaction 1 mole of A reacts with 2 moles of B to produce 3 moles of C. Now using the concept of mole one can easily measure the amount of reactants reacted and the amount of product formed, as...

                    1 Mole Exactly equals 6.022 × 10²³ particles

                    1 Mole of Gas (at STP) exactly occupies 22.4 L Volume

                    1 Mole of any compound exactly equals the molar mass in                                 grams

Therefore, <span>Stoichiometry is very helpful in quantitative analysis.</span>
6 0
3 years ago
Read 2 more answers
A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

7 0
3 years ago
How many grams of barium sulfate, BaSO 4 , be precipitated when 2.25 moles of sodium sulfate , Na 2 SO 3 , reacts with an excess
gulaghasi [49]

Answer:

525.1 g of BaSO₄ are produced.

Explanation:

The reaction of precipitation is:

Na₂SO₄ (aq) + BaCl₂ (aq) →  BaSO₄ (s) ↓  +  2NaCl (aq)

Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.

The excersise determines that the excess is the  BaCl₂.

After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.

We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g

The precipitation's equilibrium is:

SO₄⁻² (aq)  +  Ba²⁺ (aq)  ⇄  BaSO₄ (s) ↓       Kps

7 0
2 years ago
Circle the solution with the highest concentration
choli [55]

Answer:

answer is option a..............

5 0
3 years ago
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