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3 years ago

10

Is there any way the pandemic helps this society?

e="22 \times 12" alt="22 \times 12" align="absmiddle" class="latex-formula">

1 answer:

Alexxandr [17]3 years ago

7 0

Answer:

u meant the answer to this ques

Explanation:

22

×12

------

44

22 ×

----------

264

-----------

or the answer can be 22×12=264

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What 2 ways that humans have contributed to desertifaction

sdas [7]

Deforestation and inappropriate agriculture otherwise known as over farming are factors of and causes of desertification.

8 0

3 years ago

What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O

Ipatiy [6.2K]

Answer:

Mass = 13.23 g

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

O₂ : C₃H₈

5 : 1

1.5 : 1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g

6 0

3 years ago

When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol$

For the given chemical reaction:

$CaCO_3\rightarrow CaO+CO_2$

By stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of CaO

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of CaO

We know, molar mass of $CaO$ = 56 g/mol

Putting values in above equation, we get:

$\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

$\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%$

Hence, the % yield of the product is 87.05 %

6 0

3 years ago

Calculate the volume of the object immersed in the water in the water in the adjoining figur.

kirza4 [7]

Answer:

100 cm³

Explanation:

Hi there!

Subtract:

400 cm³ - 300 cm³

= 100 cm³

Therefore, the volume of the object is 100 cm³.

I hope this helps!

5 0

3 years ago

A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in

stira [4]

Answer:

2.13 V

Explanation:

The balanced equation of the reaction his;

2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)

Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.

E°anode= -2.87 V

E°cathode= -0.74 V

E°cell= E°cathode -E°anode

E°cell= -0.74 -(-2.87)

E°cell = 2.13 V

8 0

3 years ago