which type of energy would most likely create the worst environment disaster from an accident or spill

A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

Neutralization reaction (back titration): $NaOH+HCl\rightarrow NaCl+H_{2}O$

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0

3 years ago

In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

4 years ago

What is the concentration of nitrate ions in a 0.225 msr(no3)2 solution?

12345 [234]

Answer : The concentration of nitrate ions is 0.450 m.

Solution : Given,

Molarity of $Sr(NO_3)_2$ = 0.225 m

The balanced ionic equation is,

$Sr(NO_3)_2\rightarrow Sr^{+2}+2NO^-_3$

From the reaction, we conclude that the 1 mole of $Sr(NO_3)_2$ produces 2 moles of $NO^-_3$ ions. That means the concentration of nitrate ions is twice the value of $Sr(NO_3)_2$ .

then, 0.225 m of $Sr(NO_3)_2$ gives 2 × 0.225 of $NO^-_3$ ions

Now the concentration of $NO^-_3$ ions is equal to 0.450 m.

7 0

3 years ago

Read 2 more answers

Can someone please help me with this? This is due today, please please help me!

Elanso [62]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

ATOMIC NUMBER: It is the number of protons present in the nucleus of every atom

MASS NUMBER: It is the number of nucleon. Nucleon are the number of proton and neutrons present in the nucleus.

PROTON NUMBER: Proton number equal to the atomic number.

ELECTRON NUMBER: Electron Number is equal to the atomic number.

NEUTRON NUMBER: It is Mass Number - Atomic Number.

CHARGE: It is due to the addition or removal of the electron. + charge when electron removed and - charge when electron is added.

SYMBOL: It is represented as,

$X^Z_A$

Z is Mass Number

A is Atomic Number

<h2>_____________________________________</h2><h2>Worksheet:</h2>

(I) ATOMIC NUMBER: 17

MASS NUMBER: 35.5

PROTON NUMBER: 17

ELECTRON NUMBER: 17

NEUTRON NUMBER: 18.5

CHARGE: 0

(II) ATOMIC NUMBER: 71

MASS NUMBER: 180

PROTON NUMBER: 71

ELECTRON NUMBER: 71

NEUTRON NUMBER: 109

CHARGE: 0

(III) ATOMIC NUMBER: 40

MASS NUMBER: 86

PROTON NUMBER: 40

ELECTRON NUMBER: 38

NEUTRON NUMBER: 46

CHARGE: +2

(IV) ATOMIC NUMBER: 92

MASS NUMBER: 238

PROTON NUMBER: 92

ELECTRON NUMBER: 86

NEUTRON NUMBER: 146

CHARGE: +6

(V) ATOMIC NUMBER: 82

MASS NUMBER: 206

PROTON NUMBER: 82

ELECTRON NUMBER: 78

NEUTRON NUMBER: 124

CHARGE: +4

(VI) ATOMIC NUMBER: 34

MASS NUMBER: 79

PROTON NUMBER: 34

ELECTRON NUMBER: 36

NEUTRON NUMBER: 45

CHARGE: -2

(VII) ATOMIC NUMBER: 48

MASS NUMBER: 113

PROTON NUMBER: 48

ELECTRON NUMBER: 49

NEUTRON NUMBER: 65

CHARGE: -1

(VIII) ATOMIC NUMBER: 21

MASS NUMBER: 42

PROTON NUMBER: 21

ELECTRON NUMBER: 21

NEUTRON NUMBER: 21

CHARGE: 0

(IX) ATOMIC NUMBER:

MASS NUMBER:

PROTON NUMBER:

ELECTRON NUMBER:

NEUTRON NUMBER:

CHARGE:

(X) ATOMIC NUMBER: 83

MASS NUMBER: 209

PROTON NUMBER: 83

ELECTRON NUMBER: 80

NEUTRON NUMBER: 126

CHARGE: +3

(XI) ATOMIC NUMBER: 47

MASS NUMBER: 108

PROTON NUMBER: 47

ELECTRON NUMBER: 47

NEUTRON NUMBER: 61

CHARGE: 0

(XII) ATOMIC NUMBER: 49

MASS NUMBER: 116

PROTON NUMBER: 49

ELECTRON NUMBER: 46

NEUTRON NUMBER: 67

CHARGE: +3

(XIII) ATOMIC NUMBER: 53

MASS NUMBER: 128

PROTON NUMBER: 53

ELECTRON NUMBER: 54

NEUTRON NUMBER: 75

CHARGE: -1

(XIV) ATOMIC NUMBER: 76

MASS NUMBER: 188

PROTON NUMBER: 76

ELECTRON NUMBER: 72

NEUTRON NUMBER: 112

CHARGE: +4

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />

4 0

3 years ago

Label the signals due to ha, hb, and hc in the 1h nmr spectra.

trapecia [35]

Structure along with ¹H-NMR is shown below.

Signal for Hₐ;
Based on multiplicity of of the peak, a Singlet peak (the only singlet peak present in spectrum) at 2.2 ppm was assigned to Hₐ. As the methyl group in not coupling with any other proton so it gave a singlet peak.

Signal for Hb;
Based on multiplicity of of the peak, a Quartet peak (the only quartet peak present in spectrum) at 2.4 ppm was assigned to Hb. As the methylene group in coupling with methyl group having three protons so it gave a Quartet peak.

Signal for Hc;
Based on multiplicity of of the peak, a Triplet peak (the only triplet peak present in spectrum) at 1.1 ppm was assigned to Hc. As the methyle group in coupling with methylene group having two protons so it gave a Triplet peak.

3 0

4 years ago