Look at the wave in the picture. If it has a wavelength of 2 cm, how far apart are points A and B? How do you know?

How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20

USPshnik [31]

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and the last O atoms

Reaction

Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O

1. gives a coefficient

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O

2. make an equation

Zn : left = a, right =1 ⇒a=1

H : left = b, right = 2d⇒ b=2d (eq 1)

N : left = b, right = 2+c⇒b=2+c (eq 2)

O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)

From eq 1 and eq 3

3(2d)=6+c+d

6d=6+c+d

5d=6+c (eq 4)

From eq 2 and eq 3

3(2+c)=6+c+d

6+3c=6+c+d

2c=d (eq 5)

From eq 4 and eq 5

5(2c)=6+c

10c=6+c

9c=6

c = 2/3

input eq 5

d = 2 x 2/3

d = 4/3

input eq 1

b = 2 x 4/3

b = 8/3

The equation

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to

Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

8 0

3 years ago

Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro

attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the <u>chemical reaction</u>:

$NaN_3~->~Na~+~N_2$

The next step is to <u>balance the reaction</u>:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the <u>mol calculation</u> using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the<u> molar ratio</u> between $NaN_3$ and $N_2$ we can <u>calculate the moles</u> of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can <u>calculate the grams</u>:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

I hope it helps!

5 0

3 years ago

What information do we need to determine the molecular formula of a compound from the empirical formula?

defon

Answer:

Molecular mass

Explanation:

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Thus if the molecular mass is known, then we can find the value of n which results to molecular formula.

6 0

3 years ago

What electrostatic attraction between____ forms an ionic bond?

lina2011 [118]

Between atoms (one metall and one non metall) form an ionic bond(NaCl)

4 0

3 years ago

a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0

3 years ago