__C8H18(I) + __ O2(g) —> __Co2(g) + __H2O(g) Balance out the equation

A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°C to 125°C. What is the specific heat of al

ra1l [238]

<h3>Answer:</h3>

0.90J/g°C

<h3>Explanation:</h3>

We are given:

Mass of Aluminium = 10 g

Quantity of heat = 677 Joules

Change in temperature = 125°C - 50°C

= 75°C

We are required to calculate the specific heat capacity of Aluminium

But, Quantity of heat = Mass × specific heat × Change in temperature

Q = mcΔt

Rearranging the formula;

c = Q ÷ mΔt

= 677 J ÷ (10 g × 75°C)

= 677 J ÷ 750g°C

= 0.903 J/g°C

= 0.90J/g°C

Thus, the specific heat capacity of Aluminium is 0.90J/g°C

8 0

3 years ago

A substance that is ______ will NOT dissolve in a solvent. A) freezing B) insoluble C) evaporating D) soluble

Alexxandr [17]

Answer:

insoluble

Explanation:

4 0

3 years ago

Read 2 more answers

Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y

olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$A=A_o\times e^{-\lambda t}$

Where:

$\lambda$ = decay constant

$A_o$ =concentration left after time t

$t_{\frac{1}{2}}$ = Half life of the sample

Half life of Pu-239 = $t_{\frac{1}{2}}=24,000 y$ [

$\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]$

Let us say amount present of Pu-239 today = $A_o=x$

A = ?

$A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}$

$A=0.9715\times x$

$\frac{A}{A_0}=\frac{A}{x}=0.9715$

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0

3 years ago

In this reaction, _____.

ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants) (Products)

All the statements are true.

7 0

3 years ago

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0

3 years ago