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3 years ago

Could someone please help on just question 2?

Chemistry

1 answer:

sp2606 [1]3 years ago

4 0

Hey,

A compound is a substance that is made of two or more type of atoms chemically bonded together.
So this means that they are joined together by a chemical reaction.
And of those options PF3 H2SO4 H2O And CO are compounds as you can see they are made of two different type of atoms

Hope this helps !!

You might be interested in

Can you use a meat thermometer to take your temperature?

TEA [102]

Answer:

It won't work as well. You can try though, it won't hurt anything

7 0

2 years ago

If 3.0 g of NH3 reacts with 5.0 g of HCl, what is the limiting reactant?

ELEN [110]

<h2>Answer:HCl</h2>

Explanation:

$\text{number of moles}=\frac{\text{given mass}}{\text{molar mass}}$

For $NH_{3},$

$\text{given mass}=3g$

$\text{molar mass}=14+3=17g$

$n_{NH_{3}}=\frac{3}{17}=0.176$

For $HCl,$

$\text{given mass}=5g$

$\text{molar mass}=1+35.5=36.5g$

$n_{HCl}=\frac{5}{36.5}=0.136$

The reaction between $NH_{3}$ and $HCl$ is

$NH_{3}+HCl$ → $NH_{4}Cl$

So,one mole of $NH_{3}$ requires one mole of $HCl$

$0.176$ moles of $NH_{3}$ requires $0.176$ moles of $HCl$

But there are only $0.136$ moles of $HCl$ available.

$HCl$ will be consumed first.

So, $HCl$ is the limiting reagent.

7 0

2 years ago

How many moles of HCl are present in 40.0 mL of a 0.035 M solution? 1. 0.0014 mol 2. 0.0060 mol 3. 0.25 mol?

Vinil7 [7]

40.0mL(1 L/1000 mL) = 0.040 L

then plug into the formula M = moles/liters 

0.035 M = moles/ 0.040L 

multipy both sides by 0.040L, and you get 0.0014 moles 

so the answer is 1

4 0

3 years ago

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

Answer: The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago