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JulsSmile [24]
2 years ago
15

Balance the chemical equation _CrO3->_Cr2O3+_O2

Chemistry
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

2CrO3-->2Cr2O3--->3O2

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if 11.74 liters of gas at STP are pumped into a basketball, how many moles of gas are in the basketball? assume the basketball w
ser-zykov [4K]

Answer:

0.52 mol

Explanation:

Using the general gas equation formula:

PV = nRT

Where;

P = pressure (atm)

V = volume (Liters)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At STP (standard temperature and pressure), temperature of a gas is 273K, while its pressure is 1 atm

Using PV = nRT

n = PV/RT

n = (1 × 11.74) ÷ (0.0821 × 273)

n = 11.74 ÷ 22.41

n = 0.52 mol

There are 0.52 moles in the basketball

6 0
3 years ago
What is the compound name for the formula [Ru(en)2Cl2]2+ and [Co(en)Cl2Br]-
mixer [17]

Answer:2C12

Explanation:because if u divide the the equation by 2 u well get the same answer you can multiply but that will take longer with the invisible zeros and stuff like that but the answer is 2c12 and on the real FSA u can pull out ur phone and cheat just make sure the teacher is not watching just playing with you DONT DO THAT YOU WILL GET IN SERIOUS TROBLE KIDS DO NOT DO THIS KN THE REAL DAY OF THE FSA PLEASE DONT DO IT YOU WILL GET IN SERIOUS TROUBLE WELL THATS IT FOR TODAY SO HAVE A GREAT DAY THIS IS TO PUT PRESSURE ON YOU

3 0
3 years ago
6) A volume of 473 mL of oxygen was collected at 27°C. What volume would the oxygen occupy at
sertanlavr [38]

Answer : The volume of oxygen occupy at 173° would be, 703.2 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=473mL\\T_1=27^oC=(27+273)K=300K\\V_2=?\\T_2=173^oC=(173+273)K=446K

Now put all the given values in above equation, we get:

\frac{473mL}{300K}=\frac{V_2}{446K}\\\\V_2=703.2mL

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL

6 0
3 years ago
Which of the following procedures is used during the fractional distillation of petroleum?
dem82 [27]
Increase at the temperature
7 0
3 years ago
The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.
kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0
3 years ago
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