Help Anyone please PLEASE PLEASE PLEASE ALSO PLEASE DO NOT DO THIS FOR POINTS

Can someone help, please ??

Bumek [7]

Concentrated I believe

8 0

3 years ago

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What are the three sub-types of convergent plate boundaries

Vadim26 [7]

<span>Oceanic-oceanic convergent plate boundary. 2 Continental convergent plate boundary. 3 Oceanic-continental convergent plate boundary.</span>

3 0

4 years ago

One mole of ANY element contains the:

VMariaS [17]

Answer:

1)The molar mass of an atom is simply the mass of one mole of identical atoms. However, most of the chemical elements are found on earth not as one isotope but as a mixture of isotopes, so the atoms of one element do not all have the same mass.

2)Equally important is the fact that one mole of a substance has a mass in grams numerically equal to the formula weight of that substance. Thus, one mole of an element has a mass in grams equal to the atomic weight of that element and contains 6.02 X 1023 atoms of the element.

3 0

3 years ago

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years

Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually, radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

$\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g$

8 0

4 years ago