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Luda [366]
3 years ago
12

If a black ball is denser than a white ball of the same size, what does the black ball have??

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer: The black ball will have more mass than the white ball.

Explanation: Density of a substance is defined as the ratio of mass and volume occupied by the substance.

Mathematically,

Density=\frac{Mass}{Volume}

We are given that the two balls are of same size, which means that the volume of both the balls are same. We are also given that the black ball is more dense than the white ball, which means that the black ball will have more mass.

As, Density\propto Mass

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2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.
Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance.  The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the  t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene.  so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

To learn more about  t-butoxide here

brainly.com/question/12303978

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4 0
1 year ago
What is the correct equilibrium constant
Ksju [112]

Answer:

(CO2)+(CF)/ (COF)².

Explanation:

k=

6 0
3 years ago
How do the metals in group 1 compare with the transition metals?
Nimfa-mama [501]

Answer:

Group 1 metals and transition metals are different from each other, mainly based on the colour of the chemical compounds that they form. The key difference between group 1 metals and transition metals is that the group 1 metals form colourless compounds, whereas the transition metals form colourful compounds.

6 0
3 years ago
Read 2 more answers
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
HELP ASAP I WILL GIVE BRAINLIEST
alina1380 [7]

Answer:

it can last for 30 minutes

Explanation:

because it is very good at giving off heat, extothermal heat can last for quite a while.

3 0
3 years ago
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