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polet [3.4K]
3 years ago
7

Please help like now please

Chemistry
1 answer:
DENIUS [597]3 years ago
7 0
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What type of forces allow for an object to stay at rest or continue at constant velocity.
galina1969 [7]

Answer:Inertia is the property of a body to remain at rest or to remain in motion with constant velocity. Some objects have more inertia than others because the inertia of an object is equivalent to its mass.

Explanation:

5 0
3 years ago
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How many molecules of CO2 will be produced by the decomposition
34kurt

Answer:

2.99×10²⁵ molecules of CO₂ are produced

Explanation:

Decomposition reaction is:

Ca(HCO₃)₂ => CaO(s) + 2CO₂(g) + H₂O(g)

Ratio is 1:2. Let's make a rule of three:

1 mol of bicarbonate can produce 2 moles of CO₂

Therefore, 24.9 moles of bicarbonate may produce, 49.8 moles (24.9 .2 )/1

Let's determine the number of molecules

1 mol has 6.02×10²³ molecules

49.8 moles must have (49.8 . 6.02×10²³) / 1 = 2.99×10²⁵ molecules

4 0
3 years ago
Is c10H8 a conductor
Lunna [17]

Answer:

No

Explanation:

Napthalene cannot conduct electricity

5 0
3 years ago
Write uses of Co-60 .​
ivolga24 [154]

Answer:

Cobalt Sources

Cobalt-60 is used as a radiation source in many common industrial applications, such as in leveling devices and thickness gauges. It is also used for radiation therapy in hospitals. Accidental exposures may occur as the result of loss or improper disposal of medical and industrial radiation sources.

Explanation:

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3 years ago
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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
4 years ago
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