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yawa3891 [41]
3 years ago
8

Answers to student exploration : stoichometry (GIZMOS) WORTH 100 POINTS AND BRAINLIEST

Chemistry
1 answer:
Nina [5.8K]3 years ago
5 0
I hate stoichiometry. I will help you but I need to know the question first.
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What type of chemical reaction is AgNO3(aq)+KCL(aq) AgCL(s)+KNO3(aq)​
emmasim [6.3K]

Answer:

double replacement reaction or double displacement reaction,

Explanation:

double replacement reaction, double displacement reaction, is a chemical process involving the exchange of bonds between two non-reacting chemical species which results in the creation of products with similar or identical bonding affiliations

Classically, these reactions result in the precipitation of one product.

in thía case it is AgCl

3 0
3 years ago
Can someone help its confusing
marshall27 [118]

Answer:

It's a metaphor. It's comparing Jordan and their emotions to a tornado

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3 years ago
Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil
quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

7 0
3 years ago
True are false Selective breeding is a type of genetic engineering.
musickatia [10]

Answer:

true

Explanation:

it always an enginering

8 0
3 years ago
Read 2 more answers
The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
worty [1.4K]
<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

4 0
3 years ago
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