Can a book sitting on a desk represent each of Newton's Laws?

Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density

Dmitry_Shevchenko [17]

The molality of the solution is 0.00037 m.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute in a solution.

We have the following information;

Molarity = 0.335 M

Density = 1.0432 g/mL

Temperature = 20 o C

The molality of the solution is obtained from;

m = 0.335 M × 1.0432 g/mL/ 1000(1.0432 g/mL) - 0.335 M (342 g/mol)

m = 0.344/1043.2 - 114.57

m = 0.344/928.63

m = 0.00037 m

Learn more about molality of solution: brainly.com/question/4580605

5 0

2 years ago

What happens in the process of beta decay?

Andrej [43]

Answer:

A neutron transforms into a proton and an electron.

Explanation:

i took the test got an 100%

6 0

3 years ago

One solution turns blue. A possible hydrogen ion

aalyn [17]

Answer:

1x10^-8 M

Explanation:

Since the solution turns blue, it mean the solution is a base.

Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:

1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-2

pH = 2

2. Concentration of Hydrogen ion, [H+] = 5x10-2 M

pH =..?

pH = - log [H+]

pH = - log 5x10^-2

pH = 1.3

3. Concentration of Hydrogen ion, [H+] = 5x10 M

pH =..?

pH = - log [H+]

pH = - log 5x10

pH = - 1.7

4. Concentration of Hydrogen ion, [H+] = 1x10-8 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-8

pH = 8

A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.

From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.

5 0

3 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0

3 years ago