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4 years ago

Please help!!! Its not chemistry is science.

Chemistry

1 answer:

laiz [17]4 years ago

7 0

You would want to know everything just Incase anything happens because maybe you get lost

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Hey can u answer this pls

Dmitry_Shevchenko [17]

The answer to your question is DNA!!!!! :)

5 0

3 years ago

Read 2 more answers

If a bond is currently trading at its face (par) value, then it must be the case that: A. the bond's yield to maturity is equ

Natasha2012 [34]

Answer:

A. the bond's yield to maturity is equal to its coupon rate.

Explanation:

The coupon rate is that amount of bond measured on the loan amount whereas the maturity yield is the rate held up to the settlement date and the investment arise up to the maturity date.

Since the bond is currently trading at its face value that means the bond price is equal to the face or par value

So, the appropriate option is A.

5 0

3 years ago

Read 2 more answers

Brainliest will be given to correct answer :)

Lady_Fox [76]

Answer: $\Delta G$ for the reaction is -90kJ

Explanation:

The balanced chemical reaction is,

$2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)$

The expression for Gibbs free energy change is,

$\Delta G=[n\times G_{products}]-[n\times G_{reactants}]$

Putting the values we get :

$\Delta G=[3\times G_f{S,rhombic}+2\times G_f{H_2O}]-[2\times G_f{H_2S}+1\times G_f{SO_2}]$

$\Delta G=[(3\times 0kJ/mol)+(2\times -229kJ/mol)]-[(2\times -34kJ/mol) +(1\times -300kJ/mol)]$

$\Delta G=-90kJ$

Thus $\Delta G$ for the reaction $2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)$ is -90kJ

5 0

3 years ago

Please answer right this us due today please don't guess pls

Paraphin [41]

1 is C and 2 is B :)

3 0

3 years ago

Read 2 more answers

The neutralization of H3PO4 with KOH is exothermic. H3PO4(aq)+3KOH(aq)⟶3H2O(l)+K3PO4(aq)+173.2 kJ If 60.0 mL of 0.200 M H3PO4 is

nordsb [41]

Answer:

Final temperature of solution is 27.48 $^{0}\textrm{C}$

Explanation:

Total volume of mixture = (60.0+60.0) mL = 120.0 mL

We know, density = (mass)/(volume)

So mass of mixture = $(120.0\times 1.13)g=135.6 g$

Amount of heat released per mol of $H_{3}PO_{4}$ = $\frac{(m_{mixture}\times C_{mixture}\times \Delta T_{mixture})}{n_{H_{3}PO_{4}}}$

Where, m represents mass , C represents specific heat, $\Delta T$ represents change in temperature and n is number of moles

As this reaction is an exothermic reaction therefore temperature of mixture will be higher than it's initial temperature.

Let's say final temperature of mixture is T $^{0}\textrm{C}$

So, $\Delta T_{mixture}=(T-23.43)^{0}\textrm{C}$

Here $m_{mixture}=135.6 g$ and $C_{mixture}=3.78J/(g.^{0}\textrm{C})$

Moles of H_{3}PO_{4} are added = $\frac{0.200}{1000}\times 60.0moles$ = 0.012 moles

So, $(173.2\times 10^{3})J=\frac{[(135.6g)\times (3.78J.g^{-1}.^{0}\textrm{C}^{-1})\times (T-23.43)^{0}\textrm{C}]}{0.012}$

or, T = 27.48 $^{0}\textrm{C}$

So, final temperature of solution is 27.48 $^{0}\textrm{C}$

3 0

3 years ago