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3 years ago

The equation, 2 H2(g) + O2(g) Imported Asset 2 H2O(l), represents an equation for a standard formation reaction.

Chemistry

2 answers:

disa [49]3 years ago

6 0

Answer : The given statement is false.

Explanation :

The given chemical reaction of formation of $H_2O$ is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$

Standard formation reaction :

In the standard formation reaction, one mole of substance always produced from its element that are present in their standard states.

Thus, the standard formation reaction of $H_2O$ will be:

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$

Hence, the given statement is false.

VLD [36.1K]3 years ago

5 0

If this is asking if the equation is balanced, than the answer is true

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If you were sitting near a gram of protactinium-234 and a gram of uranium-234, both solid materials, which would you consider mo

Novosadov [1.4K]

aswer: protactinium-234 is more dang erous than uranium

6 0

3 years ago

A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.

sleet_krkn [62]

Explanation:

The given data is as follows.

Pressure (P) = 760 torr = 1 atm

Volume (V) = $720 cm^{3}$ = 0.720 L

Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

Total atoms present (n) = $\frac{PV}{RT}$

= $1 \times \frac{0.720 L}{0.0821 \times 298}$

= 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

x + y = 0.0294

Also, 40x + 131y = 2.966

x = 0.0097 mol

y = (0.0294 - 0.0097)

= 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

= $\frac{0.0197}{0.0294}$

= 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0

3 years ago

If I only want to make 500 mL of a 1.25% bleach solution, how many milliliters of bleach would I need? Show your work or explain

kiruha [24]

Answer:

6.25 mL

Explanation:

1.25% of 500 mL is ...

0.0125×(500 mL) = 6.25 mL

Since 1.25% of the 500 mL of solution is bleach, that's how much you need. That amount is 6.25 mL.

4 0

3 years ago

How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac

bogdanovich [222]

You need the set of reactions that goes from ammonia to nitric acid.
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1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

2) 2NO(g)+O2(g)-->2NO2(g)

3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)

State the ratio of moles of HNO3 to NH3:

4 moles of NH3 produce 4 mole of NO,

4 moles of NO produce 4 moles of NO2

4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

=> (8/3) moles HNO3 : 4 moles NH3

Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3

Use proportions:

(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

Convert moles to grams:

molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

Answer: 3213 g.
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3 0

4 years ago

Paul determines that the hydrogen ion concentration of his unknown solution is 3.60×10^-5 M. what is the pH of this solution?

vivado [14]

Answer:

Explanation:

The pH of a substance can be found by using the formula

$p H = - log[ H^{ + } ]$

where [ H+ ] is the hydrogen ion concentration of the solution

From the question

[ H + ] = 3.60 × 10^-5 M

So the pH is

$pH = - log(3.60 \times {10}^{ - 5} ) \\ =4.44369749923$

We have the final answer as

Hope this helps you

4 0

3 years ago