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3 years ago

Which of the following is most likely an indication of a chemical change?

Chemistry

1 answer:

gayaneshka [121]3 years ago

5 0

Answer:

Explanation:

Because when a pistol is fired, chemical changes happen to light the gunpowder or whatever substance to create the noise or effect.

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Magnesium oxide decomposes into magnesium and oxygen. if 12.09 g of magnesium oxide decomposes to form 7.29 g of magnesium, what

Vlada [557]

the mass of oxy gas would be 4.03 did i calculate that right?

8 0

4 years ago

A family from Texas flies to Washington, D.C. in an airplane traveling 1,362 miles in 3 hours. What is the average speed in mile

Lady bird [3.3K]

Answer:

The answer is 454

Explanation:

1362 divided by 3 = 454

4 0

3 years ago

1) 0.143 g Mg into atoms<br> 2) 0.101 kg Ti into atoms

Neporo4naja [7]

Answer:

The answer to your question is below

Explanation:

1) 0.143g of Mg into atoms

- Look for the atomic number of Magnesium in the Periodic table

Atomic number = 24.31 g

-Use the Avogadro's number to find the number of atoms

24.31g ------------------- 6.023 x 10²³ atoms

0.143 g ----------------- x

x = (0.143 x 6.023 x 10²³) / 24.31

x = 8.613 x 10²² / 24.31

x = 3.54 x 10²¹ atoms

2) 0.101 kg of Ti into atoms

-Look for the atomic number of Titanium in the Periodic table

Atomic number = 47.87 g

-Use the Avogadro's number to find the number of atoms

47.87 g --------------------- 6.023 x 10²³

101 g ---------------------- x

x = (101 x 6.023 x 10²³) / 47.87

x = 6.08x 10²⁵ / 47.87

x = 1.27 x 10²⁴ atoms

3 0

3 years ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago

The equation for the formation of ammonia from nitrogen and hydrogen is shown below. What is the formula for the equilibrium con

WINSTONCH [101]

First, make sure to balance your equation.

3H2(g) + N2(g) ⇄ 2NH3(g)

Now, you can write your Kc expression. Remember that Kc is products over reactants, and the exponent for each product or reactant is based on its coefficient.

Kc = [NH3]^2 / [H2]^3[N2]

5 0

3 years ago