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yuradex [85]
3 years ago
13

Combustion reactions are most conveniently studied _____. hints combustion reactions are most conveniently studied _____. there

is no way to study a combustion reaction conveniently using a beaker and a thermometer using constant-pressure calorimetry using constant-volume calorimetry
Chemistry
1 answer:
Luden [163]3 years ago
3 0
By using a constant-volume calorimetry or also called as bomb calorimeter the combustion are most conveniently. So base on your choices the answer is using a constant-volume calorimetry. A constant-volume calorimetry or known as bomb calorimeter is<span> used in measuring the heat of combustion of a particular reaction.</span>
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Hydrogen gas can combine with oxygen gas under certain conditions to form water. Which name is a general term for water in this
antoniya [11.8K]

Product

Explanation:

Water in this reaction is called the product of the reaction.

The product is the new kind of matter formed as a result of a chemical change. They are produced from the combination of reactants.

                          2H₂  +   O₂        →        2H₂O

                           Reactants              Products

A chemical reaction involves the combination of chemical species to form a product. The type of reaction above is a called synthesis reaction.

learn more:

Chemical equation brainly.com/question/5247791

#learnwithBrainly

8 0
3 years ago
Will bromine react with sodium and why?
Archy [21]

<span><span>When you write down the electronic configuration of bromine and sodium, you get this

Na:
Br: </span></span>

<span><span />So here we the know the valence electrons for each;</span>

<span><span>Na:  (2e)
Br:  (7e, you don't count for the d orbitals)

Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.

However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>

<span><span /></span><span><span>

</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>

<span>
</span>

4 0
3 years ago
What is the mass of calcium in 2.3•10^23 molecules of calcium phosphate
Katena32 [7]

Explanation:

N(Ca)=3×N(Ca₃(PO₄)₂)=

=3×2.3×10²³=6.9×10²³

6.02×10²³→40g

6.9×10²³→Xg

m(Ca)=X≈45.85g

3 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

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

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
3 years ago
How many moles are in 987 grams of Ra(OH)2
ra1l [238]
Molar mass Ra(OH)2 = 260 g/mol 
<span>Moles = 987 g / 260 = 3.80 moles</span>
7 0
3 years ago
Read 2 more answers
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