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3 years ago

The atomic mass of an element is equal to the number of:

Chemistry

2 answers:

Xelga [282]3 years ago

5 0

Protons plus neutrons

Yuliya22 [10]3 years ago

4 0

Answer:

the answer is B

Explanation:

ik this bc i had it on a test that was the correct answer.

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Chemistry LA

Gre4nikov [31]

Answer:

44 g/mol

Explanation:

It should be CO2 for carbon dioxide

CO2= 12+2(16)=44

6 0

3 years ago

Enter your answer in the provided box. Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use t

MakcuM [25]

Answer : The value of $\Delta H_{rxn}$ for the reaction is, -135.2 kJ

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 formation of $ClF_3$ will be,

$ClF(g)+F_2(g)\rightarrow ClF_3(l)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)$ $\Delta H_1=167.5kJ$

(2) $2F_2(g)+O_2(g)\rightarrow 2OF_2(g)$ $\Delta H_2=-43.5kJ$

(3) $2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)$ $\Delta H_3=394.4kJ$

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) $ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)$ $\Delta H_1=\frac{167.5kJ}{2}=83.75kJ$

(2) $F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)$ $\Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ$

(3) $\frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)$ $\Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ$

The expression for enthalpy of formation of $ClF_3$ will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)$

$\Delta H_{rxn}=-135.2kJ$

Therefore, the value of $\Delta H_{rxn}$ for the reaction is, -135.2 kJ

5 0

3 years ago

Question 6 the chemical formula ch2o can be classified as: empirical only. molecular only. empirical, possibly molecular. not en

frosja888 [35]

The chemical formula CH2O can be classified as Empirical formula and possibly a molecular formula. The empirical formula of a compound is the simplest formula that indicates the simplest ratio of atoms in the compound, while the molecular formula is the representation of the actual whole number ratio between the elements of the compound.

6 0

3 years ago

in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

3 years ago

A solid precipitate is formed from the reaction of lead nitrate and sodium chromate. What would be the best method to separate t

Sliva [168]

<span>The appropriate response is filtration. Filtration is any of different mechanical, physical or organic operations that different solids from liquids by including a medium through which just the liquid can pass. The liquid that goes through is known as the filtrate</span>

8 0

3 years ago

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