How many sig figs are in<br> 0.006 <br> 0.0605<br> 60.5<br> 605.5<br> 960 x 10^-3<br> 6 <br> 60

Calculate the volume of the object immersed in the water in the water in the adjoining figur.

kirza4 [7]

Answer:

100 cm³

Explanation:

Hi there!

Subtract:

400 cm³ - 300 cm³

= 100 cm³

Therefore, the volume of the object is 100 cm³.

I hope this helps!

5 0

2 years ago

4 physical properties of spoonfull of sugar

almond37 [142]

Small crystals
<span>white, brown </span>
<span>hard as in solid at room temp </span>
<span>sweet </span>

3 0

3 years ago

Read 2 more answers

Please help me. I'll give brainliest. I am stuck. True or false Q.

belka [17]

I believe your answer is true! Hope this helps!

5 0

3 years ago

Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u

Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt) (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂ (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>

3 0

2 years ago

Read 2 more answers

Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of

DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization = 35.2 kJ/mol

Changing to J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866

ln (P2/ 1 atm)= -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm x760 = 75.09 torr

7 0

2 years ago

How many sig figs are in 0.006 0.0605 60.5 605.5 960 x 10^-3 6 60