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KatRina [158]
3 years ago
15

What happens to the air as the stove heats it?

Chemistry
1 answer:
erastova [34]3 years ago
7 0

NOTHING really but as the stove is on with water boiling the air get moist

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A)<br>Name the following compounds<br>CH, -CH-CH<br>(i)<br>CH, Br<br>-​
liq [111]

Answer:

methyl, ethyl,

Explanation:

that should be the case

8 0
3 years ago
Choose the answer below that gives the correct chemical symbol and classification (metal, metalloid, or nonmetal) for the elemen
Zielflug [23.3K]

Answer: B. The element called tin is a metal with the chemical symbol Sn.

Explanation:

A is incorrect because Ni is Nickel, not Niobium.

C is incorrect because Carbon is a nonmetal and its symbol is C, not Cr.

D is incorrect because Copper's symbol is Cu, not Ce.

E is incorrect because As is the symbol for Arsenic, not Astatine.

6 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
How many moles of propane gas would be present in 11 grams<br> of the gas at standard conditions?
Akimi4 [234]
The molar mass is usually referred to with
M
, while the mass is referred to as
m
. The amount of substance is
n
. This gives you the following relationship:
=
M
=
m
n

Since you have given (C3H8)=11 g
m
(
C
3
H
8
)
=
11

g
and you already looked up (C3H8)=44.1 gmol−1
M
(
C
3
H
8
)
=
44.1

g
m
o
l
−
1
, you can use this formula to determine (C3H8)
n
(
C
3
H
8
)
.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.
5 0
3 years ago
Li + HNO3 &gt; LiNO3 + H2 how do I balance it? Show work pls
azamat

Answer: The balanced equation is 2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}.

Explanation:

The given reaction equation is as follows.

Li + HNO_{3} \rightarrow LiNO_{3} + H_{2}

Number of atoms present on reactant side are as follows.

  • Li = 1
  • H = 1
  • NO_{3} = 1

Number of atoms present on product side are as follows.

  • Li = 1
  • H = 2
  • NO_{3} = 1

To balance this equation, multiply Li by 2 and HNO_{3} by 2 on reactant side. Also, multiply LiNO_{3} by 2 on product side.

Hence, the equation can be rewritten as follows.

2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}

Now, number of atoms present on reactant side are as follows.

  • Li = 2
  • H = 2
  • NO_{3} = 2

Number of atoms present on product side are as follows.

  • Li = 2
  • H = 2
  • NO_{3} = 2

As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.

Thus, we can conclude that the balanced equation is 2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}.

3 0
2 years ago
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