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saveliy_v [14]
2 years ago
15

BRAINLIEST FOR RIGHT ANSWERS PLS HELP

Chemistry
1 answer:
AVprozaik [17]2 years ago
5 0

Answer:

Lower

Option B

Pls mark it as brainlist.

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El numero de moles en 70 gramos de HF
saveliy_v [14]

Answer:

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Explanation:

7 0
3 years ago
Read 2 more answers
The weight of the body decrease inside water why?​
Lady_Fox [76]

Penurunan atau kehilangan massa otot bisa menimbulkan penurunan berat badan yang tidak direncanakan

3 0
3 years ago
Calculate kb the base dissociation constant for [c2h3o2-], acetate anion, for each of your trials from the concentrations of spe
Pani-rosa [81]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Ka (mol/L ) = <span>0.00002340</span>

5 0
3 years ago
If you start with 3 moles of sodium and 3 moles of chlorine to produce sodium chloride, what is the limiting reagent?(you will n
Harman [31]

Sodium(Na) is the limiting reagent.

<h3>What is Limiting reagent?</h3>

The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.

How do you identify a limiting reagent?

The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.

Start by writing the balanced chemical equation that describes this reaction

2Na_{(s)} + Cl_{2 (g)} -- > 2NaCl_{(s)}

Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.

now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.

to learn more about Limiting Reagent go to - brainly.com/question/14222359

#SPJ4

4 0
1 year ago
An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

7 0
3 years ago
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