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3 years ago

how much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?

1 answer:

7 0

1 gallon of antifreeze = 60% of mixture. Total mixture:
Vm = 1/0.6
= 1.67 gallons

Volume of water = total vol - antifreeze vol
= 1.67 - 1
= 0.67 gallon of water

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Carbon-12 contains 6 protons and 6 neutrons. The radius of the nucleus is approximately 2.7 fm (femtometers) and the radius of t

Slav-nsk [51]

Explanation:

Let us assume that volumes is equal to spheres volume of nucleus.

fm = femto meter = $10^{-15}$

pm = pico meter = $10^{-12}$

Hence, calculate the volume as follows.

V = $\frac{4}{3} \times \pi \times r^{3}$

= $\frac{4}{3} \times 3.14 \times (2.7)^{3}$

= 82.4481 $fm^{3}$ (cubic femtometers)

or, = 82 $fm^{3}$

Now, we will calculate the volume of the atom as follows.

V = $\frac{4}{3} \times \pi \times r^{3}$

= $\frac{4}{3} \times 3.14 \times (70)^{3}$

= 1436758.4 $pm^{3}$ (cubic picometers)

thus, we can conclude that volume of the nucleus is 82 $fm^{3}$ and volume of the atom is 1436758.4 $pm^{3}$ .

7 0

2 years ago

Which of the following compounds has the most deshielded protons?A) CH3ClB) CH3IC) CH3BrD) CH4

Korolek [52]

Given :

Some compounds :

$A)\ CH_3Cl\ B)\ CH_3I\ C)\ CH_3Br\ D)\ CH_4$ .

To Find :

Which of the following compounds has the most deshielded protons .

Solution :

Deshielded means nucleus whose chemical shift has been increased due to removal of electron density, magnetic induction, or other effects .

In simple words deshielding means the ability to shift protons .

Now , among Cl , I , Br and H . Cl is the most electron negative .

Therefore , deshielding will be more in $CH_3Cl$ .

Hence , this is the required solution .

6 0

3 years ago

I NEED HELP PLEASE !!!!

inysia [295]

Answer:

over 10 weeks the plant will grow 24 time longer than the the same hieght of the plant and the plant age will become 45 and so the plant is very much give hey man is my answer corecct

Explanation:

3 0

2 years ago

Read 2 more answers

Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s

Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0

3 years ago

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0

2 years ago