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3 years ago

What would happen if uranium-241 lost 2 protons and 2 neutrons?

1 answer:

5 0

If uranium-241 lost 2 protons and 2 neutrons, then thorium-237 would be produced. The number beside the element is the mass number which is the sum of protons and neutrons. If uranium would lose 2 protons and 2 neutrons, then the mass number would decrease by 4 making it 237. Looking in a periodic table, thorium is the element which has a mass number of 237.

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Which option is the most accurate description of speed?

andrezito [222]

3. is the correct option listed above.

3 0

3 years ago

Read 2 more answers

Given that 10.00mL of 0.1894M CH3COOH was titrated with 0.2006M NaOH in this experiment, calculate the volume, in mL, of NaOH re

sertanlavr [38]

Answer:

$V_{base}=9.442mL$

Explanation:

Hello!

In this case, when acetic acid is titrated with sodium hydroxide, the following chemical reaction is carried out:

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

Whereas there is a 1:1 mole ratio between the acid and the base, which means that at the equivalence point we evidence:

$n_{acid}=n_{base}$

Which in terms of volumes and concentrations is written as:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, solving for the required volume of base, we obtain:

$V_{base}=\frac{M_{acid}V_{acid}}{M_{base}}=\frac{0.1894M*10.00mL}{0.2006M}\\\\ V_{base}=9.442mL$

Best regards!

7 0

3 years ago

. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in

valentina_108 [34]

Answer : The the rate expression will be:

$Rate=-\frac{d[Cl_2]}{dt}=-\frac{1}{3}\frac{d[F_2]}{dt}=+\frac{1}{2}\frac{d[ClF_3]}{dt}$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the rate expressions for this given reaction.

The given balanced equations is:

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)$