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2 years ago

Is water wet yes or no

Chemistry

2 answers:

tatyana61 [14]2 years ago

6 0

Answer:

no water isn't wet.

Explanation:

Water isn't wet by itself, but it makes other materials wet when it sticks to the surface of them

Ipatiy [6.2K]2 years ago

3 0

Maybe. Like in hitch hikers guide to the galaxy, I need to consult with the super computer

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How many milligrams are in 2.89 x 1024 CO2 molecules?

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Answer:

2,959.36

Explanation:

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2 years ago

You are a marine biologist who wants to study what type of zooplankton (microscopic animals that live in the ocean and feed larg

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3 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

A sample of gas occupies a volume of 350.0 mL at 840mm Hg and 33°C. Determine the volume of this sample at 600 mm By and 52°C

insens350 [35]

Answer:

V₂ = 520.42 mL

Explanation:

Given data:

Initial volume = 350.0 mL

Initial pressure = 840 mmHg

Initial temperature = 33°C (33 +273 = 306 K)

Final temperature = 52°C (52+273 = 325 K)

Final volume = ?

Final pressure = 600 mmHg

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 840 mmHg × 350.0 mL × 325 K / 306 K × 600 mmHg

V₂ = 95550000 mmHg.mL.K /183600 K.mmHg

V₂ = 520.42 mL

7 0

2 years ago

The half-life of strontium-90 is 28.1 years. Calculate the percent of a strontium sample left after 100 years.

Aliun [14]

Answer:

Option 4. 8.5%

Explanation:

The half-life time of a radiactive isotope (radioisotope) is a constant value, meaning that the amount of the radioisotope that decays will be (1/2) raised to the number of half-lives passed.

Naming A₀ the initial amount to the radioisotope, you can build this table to find the amount left.

Number of half-lives amount of radiosotope left

0 A₀

1 (1/2) × A₀

2 (1/2)×(1/2)×A₀ = (1/2)² × A₀

3 (1/2)³ ×A ₀

4 (1/2)⁴ × A₀

n (1/2)ⁿ × A₀

Now calculate the number of half-lives the strontium-90 sample has passed after 100 years:

n = 100 years / 28.1 years ≈ 3.5587

Hence, the amount of strontium-90 is:

$(\frac{1}{2})^{3.5587}A_0=0.08486A_0$

In percent, that is:

$(0.08486A_0/A_0).100=8.486%$

Rounding to two significant figures, that is 8.5%.

Conclusion: The percent of strontium-90 left after 100 yeaers is 8.5% (choice number 4).

6 0

3 years ago