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3 years ago

What could you do in an experiment to get more accurate results?

Chemistry

2 answers:

marissa [1.9K]3 years ago

4 0

I believe it’s b, correct me if i’m wrong

alexandr1967 [171]3 years ago

4 0

It will be D because it is always good practice to take more than five recordings/results/observations so that if you realize that the results vary a bit, you can find the average.

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alukav5142 [94]

decameters - meters: multiply by 10

meters to meters: multiply by 1

centimeters to meters: divide by 100

millimeters to meters: divide by 1000

For the rows at the bottom:

hectometer row: 100, multiply by 100, 4500

decameter row: 10, multiply by 10, 450

meter row: 1, multiply by 1, 45

decimeter row: 0.1, divide by 10, 4.5

centimeter row: 0.01, divide by 100, 0.45

im guessing theres a millimeter row at the bottom:

millimeter row: 0.001, divide by 1000, 0.045

hope this helps!

5 0

3 years ago

Determine which elements or compounds are products in the<br> follow reaction equation:

Anna71 [15]

Answer: D is the answer since it is the product of this equation

Explanation: HOPE I AM RIGHT AND IT HELPS!!!

need more explanation feel free to comment in the comment box

4 0

3 years ago

One atom of silicon can properly be combined in a compound with

aev [14]

B. two atoms of oxygen

3 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$