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aleksandrvk [35]
3 years ago
6

What is beta radiation?

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

emission at a high speed or energy which often forms electrons

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What is Heterogeneous? And an example. What is Homogeneous? And an example. I just need those, thanks
Sedbober [7]
Homo means same
Hetero means different
Geneous means DNA
So homogeneous is same DNA
and heterogenous is different DNA
7 0
3 years ago
CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s
devlian [24]

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of CO will be,

CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=-110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)     \Delta H_1=110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

The expression for enthalpy change for the reaction will be,

\Delta H_{rxn}=\Delta H_1+\Delta H_2

\Delta H_{rxn}=(110.5)+(-393.5)

\Delta H_{rxn}=-283kJ/mol

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

6 0
3 years ago
What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen to make ammonia
IgorLugansk [536]
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.

We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
8 0
3 years ago
What does autosexual mean?
kolezko [41]

Answer:

Autosexuality means people are more attracted to themselves than to others and may prefer masturbation to sex with a partner.

6 0
2 years ago
Read 2 more answers
Rank the following 0.100 M solutions in order of increasing H3O+ concentration:
Inessa05 [86]

Answer:

HCN < HOCl < HF  

Explanation:

The larger the Kₐ value, the stronger the acid.

6.2 × 10⁻¹⁰ < 4.0 × 10⁻⁸ < 6.3 × 10⁻⁴

   HCN      <     HOCl    <      HF

 weakest       stronger    strongest

5 0
3 years ago
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