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densk [106]
3 years ago
11

A salt derived from a weak base and a strong acid will yield what kind of salt?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0
A salt derived from a weak base and a strong acid will yield an acid salt
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Which is the best practice recommended in the safety video to mix and acid or a base with a solvent?
Vesna [10]

Answer:

Never pour water into acid but acid into water

Explanation:

If water is poured into extremely concentrated acid/bases, the rate of volatility and exothermic reaction is too rapid and might cause a chemical eruption, leading to acid burns.

Safety precautions hence dictate the reverse is practiced.

I believe this is a clear answer.

4 0
3 years ago
PLEASE I NEED HELP ASAP!!!!
miv72 [106K]

The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of  CaCO3 reacts with  0.4 moles  * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2  produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 --->  Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol =  0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles  reacts with   0.25 moles   * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 =  0.04 moles of CO2 * 74 g/mol = 2.96  g

Percent yield = 2.5 g/ 2.96  g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

#SPJ1

5 0
1 year ago
Pls help! Polonium has a large, unstable nucleus. Through which process is it most likely to become stable?
Stella [2.4K]

The process through which Polonium is most likely to become stable is: B. alpha decay.

An unstable element refers to a chemical element that lose particles because its nucleus contain an excess of internal energy (neutron or proton).

This ultimately implies that, an unstable element is radioactive in nature.

In Science, some examples of an unstable element are:

  • Tritium.
  • Bismuth-209 .
  • Xenon.
  • Polonium.

Polonium is a chemical element with a large, unstable nucleus.

Basically, the most stable isotope of Polonium is Polonium-209, which typically undergoes an alpha decay to form lead-205 and the emission of an alpha particle.

⇒  ^{209}_{84}Po ----> ^{205}_{82}Pb \;+\; ^{4}_{2}\alpha

In conclusion, we can deduce from the above chemical equation that Polonium is most likely to become stable through an alpha decay.

Read more: brainly.com/question/18214726

5 0
3 years ago
Read 2 more answers
If you burn 29.4 g of hydrogen and produce 263 g of water, how much oxygen reacted?
vovikov84 [41]

The reaction between hydrogen and oxygen to form water is given as:

H_2 + O_2 \rightarrow H_2O

The balanced reaction is:

2H_2 + O_2 \rightarrow 2H_2O

According to the balanced reaction,

4 g of hydrogen (4\times 1) reacts with 32 g of oxygen (2\times 16).

So, oxygen reacted with 29.4 g of hydrogen is:

\frac{29.4\times 32}{4} = 235.2 g

Hence, the mass of oxygen that is reacted with 29.4 g of hydrogen is 235.2 g.

7 0
3 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
3 years ago
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