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3 years ago

15

A substance is tested with litmus paper. the blue litmus paper remains blue while the red limtus paper remains red. what can you

conclude about the substance

1 answer:

stich3 [128]3 years ago

4 0

It’s neutral with a pH of 7. It’s probably water.

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What would the result be of adding one proton to an atom

GREYUIT [131]

Answer:Protons carry a positive electrical charge and they alone determine the charge of the nucleus. Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium and thats your answer

Explanation:

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3 years ago

Help with atoms WILL MARK BRAINLIEST

Eddi Din [679]

Answer:

C 1:1

Explanation:

Hydrogen loses an electron to becone +1 which is a cation and flourine gains that electron to have a full outer shell on the 2p subshell to become -1 and is an anion so the ratio is 1:1.

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2 years ago

Which of these statements about the Plum-Pudding Model is incorrect?

Licemer1 [7]

Answer:

It is a model of the atom that has electrons surrounding the nucleus

Step by step explanation:

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3 years ago

A catalyst is used to increase the reaction rate of a chemical reaction.

hichkok12 [17]

The answer would be C

3 0

3 years ago

If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°

Likurg_2 [28]

The final temperature of the mixture : 21.1° C

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in(gained) = Q out(lost)

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Q ethanol=Q water

mass ethanol=

$\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g$

mass water =

$\tt mass=1~g/ml\times 45~ml=45~g$

then the heat transfer :

$\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC$

5 0

3 years ago