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Digiron [165]
3 years ago
6

Help please!!!!!!!!! :(

Chemistry
2 answers:
vladimir2022 [97]3 years ago
8 0

Answer:1 and 4

Explanation:

take notes

harkovskaia [24]3 years ago
8 0
The answer is response number A
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A cylinder with a radius of 5.6 cm and a height of 10.7 cm has what volume? (You can look up the formula for the volume of a cyl
Keith_Richards [23]

Answer:

V=1054.2cm^{3}

Explanation:

1. First take the cylinder volume formula:

V=\pi.r^{2}.h

2. Then take the values for the radius and the height of the cylinder and replace them into the formula:

V=\pi*(5.6cm)^{2}*(10.7cm)

3. Solve the equation:

V=\pi*(31.36cm^{2})*(10.7cm)

V=\pi*335.5cm^{3}

V=1054.2cm^{3}

6 0
4 years ago
What indicates that two objects are in thermal equilibrium?
Leona [35]

Answer:

B

Explanation:

equilibrium happens when two things are in complete balance and thermal is in reference to heat so two things are in thermal equilibrium once they have the same temperature.

4 0
2 years ago
What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 11.630 g to 4.000 g ?
Dima020 [189]

The amount of sugar is 2621 mg

Why?

The complete question is:

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.

Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?

To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:

\frac{8.315 gSugar}{12.630 g Chocolate}*\frac{1000mg}{1g}=658.35mgSugar/gChocolate

Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:

4.000gChocolate*\frac{658.35mgSugar}{1gChocolate}=2621mgSugar

Have a nice day!

5 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
4 years ago
What would I write in there? I’m confused
pav-90 [236]
I assume what they are asking you? Sorry if that sound mean
3 0
3 years ago
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