All categories

3 years ago

What is the energy change when the temperature of 14.8 grams of solid lead is decreased from 38.4 °C to 23.7 °C ?

Chemistry

1 answer:

zimovet [89]3 years ago

8 0

Answer: Q= -34.6 J

Explanation: Use Q= mc∆T

c for solid lead is equal to 0.159 J/g°C

Substitute the values:

=14.8 g(0.159J/g°C)(23.7°C-38.4 °C)

= 14.8 g(0.159J/g°C) ( -14.7°C)

= -34.6 J

Cancel out the following units: g and °C so the remaining unit is in J.

You might be interested in

The formula K2S indicates that:

sattari [20]

The answer is C. there are two potassium ions for each ion of sulfur

4 0

3 years ago

145.8g of Mg contains how many moles?

Tems11 [23]

A: 6 moles
You have to find the molar mass and then multiply

8 0

3 years ago

What is the identity of the atom shown?

posledela

Answer:

Explanation:

Argon has 17 protons and 18 neutron

atomic number is the same as the protons number.

3 0

3 years ago

Read 2 more answers

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

Find the mass of 3.00 mol of acetic acid, C2H4O2.

Alexeev081 [22]

Molar mass

C₂H₄O₂ = 60.0 g/mol

n = mass / molar mass

3.00 = mass / 60.0

m = 3.00 * 60.0

m = 180 g of C₂H₄O₂

hope this helps!

5 0

3 years ago