Is heating sulfur and copper a physical or chemical change?

What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3

Allisa [31]

O.N. of Na = +1

O.N. of O = -2

Let, O.N. of Tin = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2

SO OPTION C IS YOUR ANSWER......

7 0

3 years ago

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What is the name of the binary compound that has the formula k2s?

kozerog [31]

Potassium sulfide, also known as dipotassium monosulfide, consists of two potassium ions bonded to a sulfide atom, rendering the chemical formula K2S.<span>Rarely found in nature due to its high reactivity with water, potassium sulfide is refined from the more common potassium sulfate (K2SO4) and is used in many industries</span>

8 0

3 years ago

2.

katovenus [111]

Answer:

A. 6atm

Explanation:

Using pressure law equation:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

According to this question;

P1 = 3 atm

P2 = ?

T1 = 120K

T2 = 240K

Using P1/T1 = P2/T2

3/120 = P2/240

Cross multiply

240 × 3 = P2 × 120

720 = 120P2

P2 = 720/120

P2 = 6atm

4 0

3 years ago

If there are 80 liters of argon gas in a piston and the piston expands the gas until its volume is 2000 liters with a pressure o

Mice21 [21]

Answer: The correct answer is A. 11.5 atm. The temperature is held constant at 293 K, therefore, we can use Boyle's Law to determine the initial pressure. Boyle's Law states that there is an inverse relationship between pressure and volume of gases. Therefore, as volume increases, the pressure will decrease and vice versa.

Further Explanation:

Boyle's Law can be mathematically expressed as:

$P_{initial}V_{initial}\ = P_{final}V_{final}$

In this problem, we are given the values:

P(initial) = ?

V(initial) = 80 L

P (final) = 0.46 atm

V (final) = 2000 L

Plugging in these values into the equation:

$P_{initial}(80 \ L) = \ (0.46 \ atm)(2000 \ L)\\P_{initial} \ = \frac{(0.46 \ atm) \ (2000 \ L)}{80 \ L}\\P_{initial}\ = 11.5 atm\\$

The initial pressure was 11.5 atm. Since the volume increased or expanded, the space where the gas particles move is bigger, so the frequency of collisions with the wall of the container and with other particles are effectively decreased. This, therefore, decreases the pressure from 11.5 to 0.46 atm.

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