Can someone please help me correct A and B please!!!!!!!!

Chemistry

1 answer:

ale4655 [162]3 years ago

6 0

A) a good hypothesis to be put in question for the experiment would be “Fluorescent light can fade color photographs quicker than direct sunlight can.”

B) the controlled variables, or things that will need to be kept constant throughout the experiment, would be the amount of light on each picture, the amount of ink on each picture, the type of ink on each picture, the size of each picture, and the time of day of the light for each picture.

You might be interested in

Traits that are due to _______ are considered to be acquired traits.

kirill [66]

Environmental conditions

Explanation:

Traits that are due to environmental conditions are considered to be acquired traits. This kind of trait are not coded in the DNA and passed down as inherited traits.

A larger body muscle due to regular exercise and fitness training is an acquired trait.
Skin color most times are inherited.
Acquired traits are typically conditioned due to the environment an organism finds itself.
These traits cannot be passed down.

Learn more:

Traits brainly.com/question/2908634

#learnwithBrainly

6 0

3 years ago

What is metallic conductivity ?

melamori03 [73]

Explanation:

metallic conduction:the movement of electrons through a metal, with no changes in the metal and no movement of the metal atoms.

I hope this help uuhh............

6 0

2 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .