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3 years ago

10) What are the four main spheres of planet?

Chemistry

1 answer:

Tomtit [17]3 years ago

5 0

Answer:

the four main spheres of the earth are geosphere, hydrosphere, atmosphere and biosphere

Explanation:

geosphere consists of all rocks on Earth

atmosphere which are the gases that surrounds the earth

hydrosphere which is all the water on the earth

biosphere which are the living things on the earth

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A hydrogen filled balloon has a volume of 8.3 L at 36°C and 751 mmHg. How many moles of hydrogen are inside the balloon?

tamaranim1 [39]

Answer: I think the formula is PV=nRT and I divide both sides by RT, but this is as far as I can get in my equation before I get stumped: (751 mm Hg) (8.3 L)/ (309 K) Can you help?

Explanation:

8 0

3 years ago

Draw a line graph showing the population of beetles and healthy trees in the forest for five years. Round the numbers to the nea

Sergio [31]

how do i draw a line ghraph on a computer

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3 years ago

Read 2 more answers

A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25Â°C. Calculate the p

natima [27]

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of $HC_2H_5CO_2$ = 0.1917 M

Volume of $HC_2H_5CO_2$ = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of $HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2$

$\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422$

As we known that at equivalent point, the moles of $HC_2H_5CO_2$ and KOH are equal.

So, Moles of KOH = Moles of $HC_2H_5CO_2$ = 0.0422 mol

Now we have to calculate the volume of KOH.

$\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}$

$\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}$

$\text{Volume of }KOH=0.00754$

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

$HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O$

Moles of $C_2H_5CO_2K$ = 0.0422 mol

$\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M$

At equivalent point,

$pH=\frac{1}{2}[pK_w+pK_a+\log C]$

Given:

$pK_w=14\\\\pK_a=4.89\\\\C=0.1855M$

Now put all the given values in the above expression, we get:

$pH=\frac{1}{2}[14+4.89+\log (0.1855)]$

$pH=9.08$

Therefore, the pH at equivalence is, 9.08

6 0

3 years ago

4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/

nalin [4]

Answer:

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.

Explanation:

Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.

Mass = Density × Volume

Mass of t-butyl alcohol = $0.79 g/mL\times 10.4 mL=8.219 g$

Moles of t-butyl alcohol = $\frac{8.219 g}{74.12 g/mol}=0.11084 mol$

Moles of 1,4-dimethoxybenzene = $\frac{5.6 g}{138.17 g/mol}=0.04052 mol$

According to reaction 2 mol of t-butyl alcohol reacts with 1 mol of 1,4-dimethoxybenzene.

Then 0.11084 moles of t-butyl alcohol will react with :

$\frac{1}{2}\times 0.11084 mole=0.05542 mol$ of 1,4-dimethoxybenzene.

This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.

According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of 1,4-di-t-butyl-2,5-dimethoxybenzene.

Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:

$\frac{1}{1}\times 0.04052 mol= 0.04052 mol$ of 1,4-di-t-butyl-2,5-dimethoxybenzene.

Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:

0.04052 mol × 250.37 g/mol = 10.144 g

Percentage yield:

$\frac{Experimental}{Theoretical}\times 100$

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:

Experimental yield = 4.2 g

Theoretical yield = 10.144 g

$\frac{4.2 g}{10.144 g}\times 100=41.40\%$

4 0

3 years ago

What is the specific activity (in Ci/g) if 1.65 mg of an isotope emits 1.56X10⁶α a particles per second?

marta [7]

The specific activity (in Ci/g) if 1.65 mg of an isotope emits 1.56X10⁶α a particles per second.

<h3>What is isotope?</h3>

Isotopes are two or more types of atoms with the same atomic number (number of protons in their nuclei) and periodic table position (and so belong to the same chemical element), but distinct nucleon numbers due to differing numbers of neutrons in their nuclei. While all isotopes of a given element have nearly identical chemical properties, their atomic weights and physical properties differ.

The term isotope is derived from the Greek roots isos and topos, which means that different isotopes of the same element occupy the same position on the periodic table. Margaret Todd, a Scottish doctor and writer, coined the phrase in 1913 as a recommendation to British chemist Frederick Soddy.

To learn more about isotopes visit:

brainly.com/question/11680817

#SPJ4

4 0

2 years ago