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2 years ago

Please help this is due soon

Chemistry

2 answers:

denis-greek [22]2 years ago

6 0

It’s b use the formula

Lubov Fominskaja [6]2 years ago

4 0

It’s b 20g/cm3 I hope this helps

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A 1.25 × 10-4 m solution of the anti-inflammatory drug naproxen has a ph = 4.2. the ka and pka of naproxen are ________ and ____

Mandarinka [93]

Naproxen is known to be a weak acid. In order to calculate its ka and pka, use the equation of getting the ph of weak acid which is ph= -1/2 log [(Ka)(Mwa)]. The Ka value is 3.18x10^-5. The pKa can be obtained through pKa = - log Ka. The pKa is 4.5.

6 0

3 years ago

Read 2 more answers

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

3 years ago

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

bulgar [2K]

Answer: Mass of $CO_2$ produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants = Mass of $CaCO_3$ + mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$ produced in this reaction was 6.56 grams

7 0

3 years ago

Give two examples of diffusion process in plants and animals.

lisabon 2012 [21]

answer:

in plants

Transport manufactured food from the leaves to others parts of the plant

Facilitates gaseous exchange through the stomata in the leaves to other parts of the plant

in animal

Exchange of respiratory gases across respiratory services

Excretion of nitrogenous waste in some unicellular organisms

Explanation:

Hope it benefit

3 0

3 years ago

What is the lewis structure for al?

bogdanovich [222]

The abbreviation Al, with one dot on top of the abbreviation, one on the left, one on the right.

8 0

3 years ago