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MrMuchimi
3 years ago
9

A calorimeter contains 251 g of water at 24.3 C.The temperature of the water decreases to 18.5 C when 12.5 g of potassium nitrat

e is dissolved in water. Is dissolving potassium nitrate in water endothermic or exothermic
Chemistry
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

Endothermic  

Explanation:

The temperature of the water decreased.

The water lost heat.

The heat must have gone into getting the KNO₃ into solution.

So, the dissolving of KNO₃ in water is endothermic.

You might be interested in
Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Comple
AfilCa [17]

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.  

Symbol of an alpha particle is ^{4}_{2}\alpha. Whereas a neutron is represented by a symbol ^{1}_{0}n, that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

        ^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n

5 0
3 years ago
Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

3 0
3 years ago
Convert 35.0 m/s to cm/min
swat32
35 m/s = 210000 cm/min
35*600=210000
3 0
3 years ago
Please Help!!!
STatiana [176]

Answer:

single replacement

Explanation:

Step 1: Data given

single replacement = A reaction in which one element replaces a similar element in a compound. For example, a metal replaces an other metal.

The general form of a single-replacement (also called single-displacement) reaction is:

A+BC→AC+B

Decomposition = a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:

AB→A+B

Synthesis =  A reaction that occurs when one or more compounds combines to form a complex compound:

A + B → AB

Double replacement: a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds.

The general form of a double-replacement reaction is:

AB+CD→AD+BC

Combustion reaction = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve  O2  as one reactant.

The reaction Zn + 2HCl → ZnCl2 + H2

⇒ Does not involve O2 = NOT a combustion reaction

⇒ The compounds do not form a complex compound = NOT a synthesis

⇒ A compound does not break down into smaller substances = NOT a decomposition

⇒ There is a replacement between Zn and H. This is a <u>single replacement</u>, not a double replacement reaction.

6 0
3 years ago
Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca
Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

k=Aexp(-Ea/RT)\\

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}

Ea_1-Ea_2 is the  lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

\frac{k_2}{k_1} =1\times 10^5

ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ

4 0
3 years ago
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