All categories

2 years ago

If you put 1000 J of heat energy into a bar of titanium and the bar's temperature rose

Chemistry

1 answer:

melisa1 [442]2 years ago

3 0

Answer:

The mass of the bar is approximately 0.33 g.

Explanation:

q (heat in joules) = 1000 J

c (specific heat capacity) = 52 J/g°C

ΔT (change in temperature) = 80 - 22 = 58°C.

m (mass in grams) = ?

m = q / cΔT

m = 1000 J / (52 J/g°C) (58°C)

m = 0.33 g

hope this helps and is right!! p.s. i really need brainliest :)

You might be interested in

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$

Calculating the molecular formula:

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.69 moles.

For Carbon = $\frac{5.082}{1.69}=3$

For Hydrogen = $\frac{11.94}{1.69}=7.06\approx 7$

For Oxygen = $\frac{1.69}{1.69}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is $C_3H_7O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

$n=\frac{118.2g/mol}{59g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2$

Hence, the molecular formula for the given organic compound is $C_6H_{14}O_2$

8 0

3 years ago

At what points does the object accelerate please explain

nirvana33 [79]

Answer:

M=3

B=25

Equation= 3x+25

Explanation:

M is your slope and if you follow the slope- intercept formula (y=m+b) you just need to plug in the numbers.

Thus your answer:

M=3

B=25

Equation= 3x+25

8 0

2 years ago

If you burn 59.1 g of hydrogen and produce 528 g of water , how much oxygen is reacted

shtirl [24]

Hydrogen + oxygen --> water
59,1g + x = 528g
528g - 59,1 = x
x = 468,9g

8 0

3 years ago

Helpppppo meeeeee plzzzzzzzzz

antoniya [11.8K]

For 2 it’s 2ft 10in
For 5 it’s 3ft 6in
You: put how tall you are
Adult: you can either measure an adult or use this 5 ft 10in

6 0

2 years ago

How does the arrangement of elements in periods relate to electron configuration

VashaNatasha [74]

Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2

Both elements' valance electrons are present in the 1st shell

(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)

4 0

3 years ago