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3 years ago

Calculate the number of kilojoules to warm 125 grams of iron from 23.5 degrees Celcius to 78 degrees Celcius

1 answer:

3 0

Q=mcΔT

q=(125g)(0.446 J/(g x °C))(78°C - 23.5°C)

q=(125 g)(0.446 J/(g x °C))(54.5°C) (Be sure to cancel all similar units)

q= 3038.375 J

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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Scilla [17]

Answer:

The essence including its given problem is outlined in the following segment on the context..

Explanation:

The given values are:

Moles of CO₂,

x = 0.01962

Moles of water,

$\frac{y}{2} =0.01961$

$y=2\times 0.01961$

$=0.03922$

Compound's mass,

= 0.4647 g

Let the compound's formula will be:

$C_{x}H_{y}O_{z}$

Combustion's general equation will be:

⇒ $C_{x}H_{y}O_{z}+x+(\frac{y}{4}-\frac{z}{2}) O_{2}=xCO_{2}+\frac{y}{2H_{2}O}$

On putting the estimated values, we get

⇒ $12\times x=1\times y+16\times z=0.4647$

⇒ $12\times 0.01962+1\times 0.03922+16\times z=0.4647$

⇒ $0.27466+16z=0.4647$

⇒ $z=0.01187$

Now,

x : y : z = $0.01962:0.03922:0.01187$

= $\frac{0.01962}{0.0118}:\frac{0.03922}{0.0188}:\frac{0.0188}{0.0188}$

= $1.6:3.3:1.0$

= $3:6:2$

So that the empirical formula seems to be "C₃H₆O₂".

8 0

3 years ago

What causes an ionic bond to form?<br><br>Look at the picture above

Gennadij [26K]

Attraction between a cation in one element and an anion in the other element

7 0

3 years ago

Read 2 more answers

QUICK! WILL MARK BRAINLIEST

nalin [4]

Answer:

Ability to be bent = Malleability

Identity = Physical Change

Electrical Current = Conductivity

Dissolve = Solubility

Color, Phase, or Hardness = Physical Property

5 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago

Rust results from iron’s reaction to oxygen. An iron nail gains mass when it rusts. How does this reaction support the law of co

Olegator [25]

Answer:

Option (A). The mass of the rusted nail equals the mass of iron and the oxygen from the air it reacted with to form the rust.

5 0

3 years ago