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3 years ago

1. How does cellular respiration add carbon to the atmosphere?

1 answer:

8 0

Glucose and oxygen are changed into energy and carbon dioxide during cellular respiration and that’s how carbon dioxide is released in the air.

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Write the correct chemical formula for the following:

nikdorinn [45]

I believe its

A. AlF

B.2S5O

C.HBr

5 0

4 years ago

At room temperature (208C) and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Ugo [173]

Total density of filled ball with nitrogen gas: $\frac{0.12G+0.6524g}{0.560L} = 1.3792g/L$

The relationship between mass and volume can be easily determined using density; for example, the mass of a body is equal to its volume multiplied by the density (M = Vd), whereas the volume is equal to the mass divided by the density (V = M/d). The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air. Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165g/L$

Mass of the nitrogen gas : $1.165g/L \times 0.560L = 0.6524g$

Learn more about Mass and Density here:

brainly.com/question/10821730

#SPJ4

4 0

2 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

4 years ago

The decomposition of hydrogen peroxide (H2O2(aq)) is proposed to follow a reaction mechanism of: Step 1: H2O2(aq) + Br–(aq)→H2O(

lys-0071 [83]

Answer:

Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without

Explanation:

Let us consider the equation below:

Step 1:

H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)

Step 2:

BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)

From the above equation, we can see that Br– is unchanged.

This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.

8 0

3 years ago

How many protons and neutrons are in the nucleus of isotope with mass of 68.926 amu?

Sidana [21]

There are 30 protons and 39 neutrons in the nucleus.

This must me the isotope of an element with an atomic mass close to 69 u.
The only candidates are Zn and Ga.
Zn has a zinc-69 isotope with mass 68.926 u.
Ga has a gallium -69 isotope with mass 68.925 u.
The isotope is probably $_{30} ^{69}Zn$ .
It has 30 protons and 39 neutrons.

4 0

3 years ago