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3 years ago

5

True or False: When it is spring in the Northern Hemisphere, it is autumn in the Southern Hemisphere.

2 answers:

dsp733 years ago

8 0

Answer:

True

Explanation:

svlad2 [7]3 years ago

3 0

Answer:

That is true

Explanation:

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How many grams of tris (mw 121.1) would you need to prepare 100 ml of a 100 mm tris solution? _________ grams?

Tems11 [23]

Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)

Mass = Molarity x molar mass x volume of solution in Litres

Molarity of Tris = 100 mM = 0.1 M

volume of Tris sol. = 100 mL = 0.1 L

molar mass of Tris = 121.1 g/mol

Hence,

mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution

= 0.1 M x 121.1 g/mol x 0.1 L

= 1.211 g

mass of Tris = 1.211 g

7 0

3 years ago

Which best describes the transfer of heat energy through conduction? A. Heat energy is transferred from molecule to molecule by

TiliK225 [7]

It should be option (A) because conduction is the transfer of heat energy from one place to another by vibrations/motion of the molecules. I hoped I helped.

5 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago

Alchen [17]

Balanced:
1. <span>Na2O + H2O ---> 2NaOH
2. </span><span>K2O + H2O ---> 2KOH
3. </span><span>MgO + H2O ---> Mg(OH)2
4. </span><span>CaO + H2O ---> Ca(OH)2
5. </span><span>SO2 + H2O ⇄ H2SO3
6. </span>SO3 + H2O ---> H2SO4
All except by 2 were balanced.

7 0

3 years ago

If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution

Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution = $5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

( $1\mu M=10^{-6} M$ )

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$

8 0

4 years ago