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SSSSS [86.1K]
4 years ago
8

EMERGENCY HELP DUE IN 10 MIN!! I WILL MARK BRAINLIEST WHOEVER ANSWERS FIRST JUST PLS HELP!!

Chemistry
2 answers:
Airida [17]4 years ago
6 0
If the object has a lower density than the fluid it is in, it will float.

So for the wood one, water has a density of 1g/cm so the wood would float.

For the second one the vegetable oil has the lowest density since it’s at the top
Butoxors [25]4 years ago
3 0

Answer:

1 float

2 vegetable oil

Explanation:

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Why are animal vaccinated?
juin [17]

Answer:

to go against dangerous viral and bacteria

Explanation:

5 0
3 years ago
Read 2 more answers
What is the name of the compound Ca(NO3)2 ?
natulia [17]

Answer: B) calcium nitrate

5 0
3 years ago
6.201cm+7.4cm+0.68 cm+12.0cm
zimovet [89]
If you want the answer is centimeters is it going to be:
26.281 cm
If it make it easier for you to solve add 6.201 and 7.4 which will equal 13.601. Then add .68 to 13.601 which equals 14.281. Last add 12 to 14.281 which equals 26.281.

Hope this helps.

8 0
3 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
What is the enthalpy of fusion of water?
malfutka [58]

Answer:

d-  334 kJ/g.

Explanation:

You can detect it from the units of the different choices.

a- has the unit J/g.°C that is the unit of the specific heat capacity (c).

b- has the unit Kelvin that is the unit of temperature.

c- has the unit g/mol which is the unit of the molar mass.

d- has the unit kJ/g which is the unit of the enthalpy divided by the no. of rams that is the specific entha;py of fusion.

<em>So, the right choice is: d-  334 kJ/g.</em>

5 0
3 years ago
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