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2 years ago

HELP ITS SCIENCE PLS

Chemistry

2 answers:

gulaghasi [49]2 years ago

8 0

15. 7

16. 30 (but not sure)

17.Conditions such as high light intensity result in an increased stomata index

19. They grow toward the sunlight to be able to generate energy by photosynthesis

20. If you plant a seed on its side, the shoot and root will emerge horizontally, but quickly change their direction in growth.

21. When the soil of a plant runs too low of available water, the water chains in the xylem become thinner and thinner due to less water...Occasionally, vascular fungal diseases will clog the xylem tissue and cause wilting.

anygoal [31]2 years ago

6 0

Don’t know lol um maybe

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The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

5 0

3 years ago

If 1 mol of XO₂ contains the same number of atoms as 60 g of XOs, what is the molar mass of XO₂?

Lena [83]

Answer:

44 grams/mole

Explanation:

If 1 mol of XO₂ contains the same number of atoms as 60 g of XO3, what is the molar mass of XO₂?

60 grams of XO3 is one mole XO3, since it has the same number of atoms as 1 mole of XO2.

Let c be the molar mass of X. The molar mass of XO3 is comprised of:

X: c

3O: 3 x 16 = 48

Total molar mass of XO3 is = 48 + c

We know that the molar mass of XO3 = 60 g/mole, so:

48 + c = 60 g/mole

c = 12 g/mole

The molar mass of XO2 would be:

1 X = 12

2 O = 32

Molar mass = 44 grams/mole, same as carbon dioxide. Carbon's molar mass is 12 grams.

5 0

2 years ago

A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon

Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0

3 years ago

If you need 2.5 moles of Rb2Cr7, how many grams do you need?

BartSMP [9]

Answer:

Explanation:

5 0

2 years ago

It is common for students to overshoot the endpoint, meaning they add too much NaOH(aq) from the buret, which causes the solutio

umka2103 [35]

Answer: the percentage of acetic acid will be low.

Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.

Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.

This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)

8 0

3 years ago