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cricket20 [7]
2 years ago
9

How many seconds long is this chemistry class if it last 40.0 minutes

Chemistry
2 answers:
Westkost [7]2 years ago
6 0

Answer:

2,400 seconds

Explanation:

I multiplied the time value by 60 to get my answer

Hope this helped!

ziro4ka [17]2 years ago
4 0
Around 2,400 seconds would be in a 40.0 minute class
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What is the balanced equation and type of reaction for: <br> Copper metal+aqueous silver nitrate
Gnesinka [82]

Answer:

Cu(s) + 2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

This chemical equation means:

One mole of solid copper plus two moles of aqueous silver nitrate produce one mole of copper(II) nitrate plus two moles of solid silver.

This is a single replacement reaction in which the metal copper replaces the metal silver.

3 0
2 years ago
The nuclear fission process releases neutrons and
yuradex [85]
<span>the answer is c. energy</span>
3 0
3 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
Round off 4.5778 x 10mg to three significant figures.
lakkis [162]

Answer:

Explanation:

= 45.8

4 0
3 years ago
How many kilojoules are released when 116 g of Cl2 reacts with silicon? Si(s)+2Cl2(g)→SiCl4(g)ΔH=−657kJ
alexandr402 [8]

Answer:-  537 kJ of heat is released.

Solution:- For the given equation, \Delta H is -657 kJ and the coefficient of Cl_2 in the balanced equation is 2. It means 657 kJ of heat is released when 2 moles of chlorine are used. We need to calculate the heat released when 116 g of Cl_2 are used.

Grams of chlorine are converted to moles and then multiplied by the \Delta H value and divided by the coefficient of chlorine and the set could be shown using dimensional analysis as:

116gCl_2(\frac{1mol}{70.9g})(\frac{657kJ}{2mol})

= 537.46 kJ

If we use the correct sig figs then it needs to be round off to three sig figs as the given grams of chlorine has only three sig figs. So, 537 kJ of heat is released.

6 0
2 years ago
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