What other product occurs when ac-222 releases an alpha particle?

A gardener waters the plants in a garden using a garden hose during a drought. The gardener observes that most of the water give

Anna71 [15]

Answer:

Break up the soil to increase the number of pores should be your answer.

Explanation:

If you do that it will increase the amount of water in the plant.

8 0

2 years ago

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a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp

Drupady [299]

Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL$

Putting values in above equation, we get:

$1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148$

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_3PO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL$

Putting values in above equation, we get:

$3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M$

The concentration of the phosphoric acid solution is 1.172 M

3 0

2 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

Which of the following statements about metalloids is true?

Klio2033 [76]

Metalloids had properties that fall between those of metals and nonmetals (I believe that to be correct-.-)

7 0

2 years ago

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An element that can have different physical properties or chemical arrangement is called a (n)

shusha [124]

<h3>Answer:</h3>

D. Allotrope

<h3>Explanation:</h3>

What is allotropy?

Allotropy refers to the existence of an element in more than one physical forms.
Allotropes are therefore different forms of an element with different physical properties or chemical arrangements.

What are some examples of allotropes?

Examples of elements that exhibit allotropy include, sulfur and carbon.
Allotropes of carbon are diamond and graphite.
Allotropes of sulfur are monoclinic sulfur and rhombic sulfur.

3 0

2 years ago