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IRISSAK [1]
2 years ago
5

A student investigated the factors that affect math grades. Which statement from the lab report best represents a conclusion for

this investigation?

Chemistry
1 answer:
Tanya [424]2 years ago
5 0

The correct answer choice from the statement above about investigated factors from lab results is :

Students who spend less time studying after school get lower math grade

Option D is the correct answer

<h3>What is mathlab?</h3>

This is an online platform where mathematics tutors help students to solve their difficulties in mathematics.

Mathematics is one of the most important aspect of knowledge in the education.

So therefore, The correct answer choice from the statement above about investigated factors from lab results is :

Students who spend less time studying after school get lower math grade

Learn more about MathLab:

brainly.com/question/13974197

#SPJ1

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The atomic mass and abundance of Cr-50 is 49.946 amu and 4.3%. The atomic mass and abundance of Cr-52 is 51.941 amu and 83.8%. T
Leviafan [203]

Answer:

52.00 AMU

Explanation:

(49.946 * 0.043) + (51.941 * 0.838) + (52.941 * 0.095) + (53.939 * 0.024) = 51.998

Make sure to round, 52.00 AMU.

3 0
3 years ago
3. Define isobar 75 points
steposvetlana [31]

Answer:

1.

meteorology A line drawn on a map or chart connecting places of equal or constant pressure.

2.

nuclear physics Either of two nuclides of different elements having the same mass number.

3.

thermodynamics A set of points or conditions at constant pressure.

5 0
3 years ago
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
What is the precipitate for CuSO4+NaOH<br> (Balancing Equations)
jeka57 [31]

Answer:

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

Explanation:

5 0
4 years ago
chegg write a net ionic equation describing the oxidation of no2 2 to no3 2 by o2 in a basic solution.
Marina86 [1]

When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

NO2→NO^−3

NO2→NO

and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−

Now add Eqn 4 and 5 (the electrons now cancel each other):

3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+

and cancel terms that’s common to both sides:

3NO2 + H2O → NO + 2NO^−3 + 2H+

This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.

Learn more about balancing equation here:

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7 0
2 years ago
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