1. Hydrogen gas burns in air according to the equation below.

Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t

barxatty [35]

Answer: -

6

Explanation: -

The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2

We see there 3 sodium on the right side from Na3AsO3.

But there are only 1 sodium on the left from NaOH.

So we multiply NaOH by 3.

As + 3 NaOH -- > Na3AsO3 + H2

Now we see the number of Hydrogen on the left is 3.

But the number of hydrogens is 2 on the left.

So, we multiply to get both sides 6 hydrogen.

As + 6NaOH -- > Na3AsO3 + 3 H2

Rebalancing for Na,

As + 6NaOH -- > 2Na3AsO3 + 3 H2.

Finally balancing As,

2 As + 6 NaOH -- > 2Na3AsO3 + 3H2

The coefficient of the NaOH molecule in the balanced reaction is thus 6

7 0

3 years ago

Attempt 2

Assoli18 [71]

Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium

4 0

3 years ago

A metal melts at 450 degrees celsius. Is this property of the metal classified as chemical or physical?

Romashka-Z-Leto [24]

It's a physical property because it doesn't change the element of the metal.
does this make sense?

8 0

3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago

The mass of a 6.0 mL sample of kerosene is 4.92. The density of kerosene is

weqwewe [10]

4.92 grams / 6 mL = .82 grams / mL

(A) 0.82 g/mL

3 0

3 years ago