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3 years ago

Based on the information provided in the graph, we can see that as the independent variable increases,

Chemistry

2 answers:

Gnom [1K]3 years ago

4 0

Well I have no clue on what your class is doing but to me, it looks like B. the dependent variable also increases

sergejj [24]3 years ago

4 0

Answer:

B) the dependent variable also increases.

Explanation:

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A metamorphic rock can also be thought of as a rock that changes. What causes the rock to change?

Nana76 [90]

Answer:

its ethier b or c

Explanation:

5 0

2 years ago

Read 2 more answers

(02.01 MC)

Rina8888 [55]

Answer:

C. Particle size

Explanation:

The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not

4 0

2 years ago

How many millimeters are there in 5.12 x 10^5

Anton [14]

Answer:

5.12x10¹¹ millimeters

Explanation:

Milli is a prefix used in science and engineering to decribe the number as the exponent x10⁻³. In the prefix kilo, the number is at the exponent x10³.

5.12x10⁵ kilometers are:

5.12x10⁵ kilometers * (1000m / 1km) = 5.12x10⁸ meters

5.12x10² meters * (1m / 1000millimeters) = 5.12x10¹¹ millimeters

8 0

3 years ago

What can we tell from the column that an element is in?

Paul [167]

Answer:

The number of electrons in the outer energy level

Explanation:

6 0

3 years ago

A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c

Andre45 [30]

Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

$E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V$

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq)+2e^-$

Reduction half reaction: $Cu^{+}(aq)+e^-\rightarrow Cu(s)$ ( × 2)

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

$E^o_{cell}=0.52-(-0.25)=0.77V$

Hence, the standard potential of the cell is 0.77 V

5 0

3 years ago