Write the overall reaction for cellular respiration

8 gm of pure calcium is treated with 50 gram of pure Hcl to give cacl2 and h2 .which is limiting reactant?

elena-14-01-66 [18.8K]

Ca as a limiting reactant

<h3>Further explanation</h3>

Given

8 g Calcium

50 g HCl

<h3>Required</h3>

Limiting reactant

Solution

Reaction

Ca + 2HCl → CaCl₂ + H₂

mol Ca (Ar = 40 g/mol) :

= mass : Ar

= 8 g : 40 g/mol

= 0.2

mol HCl (MW= 36.5 g/mol) :

= mass : MW

= 50 g : 36.5 g/mol

= 1.37

Mol : coefficient reactants :

Ca = 0.2/1 = 0.2

HCl = 1.37/2 = 0.685

Ca as a limiting reactant(smaller ratio)

3 0

3 years ago

How do astronomers observe and learn about celestial<br> objects?

MatroZZZ [7]

Answer:

Astronomers use a number of telescopes sensitive to different parts of the electromagnetic spectrum to study objects in space. Even though all light is fundamentally the same thing, the way that astronomers observe light depends on the portion of the spectrum they wish to study.

Explanation:

8 0

3 years ago

What do ALL states of matter have in common?

VLD [36.1K]

Answer:

The common thing among the three states of matter is - They are made up of small tiny particles. They have a particular mass and can occupy space . This three states have volume in it . The atoms of this three states have force of attractions between them .

Explanation:

hope this helps <3

3 0

3 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

6 0

3 years ago

How might the biodiversity of a mowed lawn compare to that of huge weedy field?

Dennis_Churaev [7]

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

5 0

3 years ago