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Leya [2.2K]
3 years ago
5

From the attitude Douglass expresses in the passage,

Chemistry
2 answers:
Katena32 [7]3 years ago
8 0

Answer:

Thrilled

Explanation:

Took the test.

Ivanshal [37]3 years ago
8 0

Answer:

the answer is thrilled

Explanation:

You might be interested in
Write a balanced chemical equation showing how each metal oxide reacts with HCl. SrO Na2O Li2O BaO
Agata [3.3K]

Answer:

See below  

Explanation:

Each metal oxide reacts with HCl to form water and the metal chloride

\rm SrO + 2HCl \longrightarrow SrCl_{2} + H_{2}O\\\\Na_{2}O + 2HCl \longrightarrow 2NaCl + H_{2}O\\\\Li_{2}O + 2HCl \longrightarrow 2LiCl + H_{2}O\\\\BaO + 2HCl \longrightarrow BaCl_{2} + H_{2}O

6 0
3 years ago
Read 2 more answers
At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution
IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = x_{A}\times P_{A}^{0}

vapour pressure of a component (B) in solution = x_{B}\times P_{B}^{0}

Where x_{A},x_{B} are mole fraction of component A and B in solution respectively

P_{A}^{0},P_{B}^{0} are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = 0.340\times 745 torr

                                                                         = 253 torr

7 0
3 years ago
Write the balanced NET IONIC equation for the reaction that occurs when perchloric acid and calcium hypochlorite are combined. I
Elanso [62]

Answer:

H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}

Explanation:

Hello there!

In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:

2HClO_4{(aq)}+Ca(ClO)_2{(aq)}\rightarrow 2HClO{(aq)}+Ca(ClO_4)_2{(aq)}

Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:

2H^++2(ClO_4)^-+Ca^{2+}+2(ClO)^-\rightarrow 2HClO{(aq)}+Ca^{2+}+2(ClO_4)^-

Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:

2H^+(aq)+2(ClO)^-(aq)\rightarrow 2HClO{(aq)}\\\\H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}

Best regards!

3 0
2 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
How many moles of h2o are produced when using 7 moles of h2
Olenka [21]
1) Write the balaced chemical equation:

H2 + 2O2 → 2H2O

2) Infere the molar ratios:

1 mol H2 : 2 mol of water

3) Make the calculus as the direct proportion relation:

[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2

As you see you produce the double number of moles of H2O than number of moles of H2 used.

Answer: 14 moles 

6 0
3 years ago
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