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2 years ago

A 21.94-gram sample of magnesium nitrate was used to create a 2.11 M solution. What is the

Chemistry

1 answer:

Tatiana [17]2 years ago

6 0

Answer:

70.1 mL

Explanation:

First let's look at the formula for magnesium nitrate, and get the molar mass, we should end up with Mg(NO3)2 for the formula and this should have a molar mass of 148.3 g/mol.

Lets get the number of moles of the magnesium by taking the number of grams over the molar mass, (21.94 g)/(148.3 g/mol). grams cancel and we're left with approximately 0.148 moles.

Now let's plug our numbers into the molarity formula, M = n/L, this should give us 2.11 mol/L = (0.148 mol)/L, now let's solve for L, divide both sides by 0.148 which will give us 14.26 L^-1 = 1/L now we take the inverse of both sides to get 0.07012 L = L.

Now we have the liters, but the question askes for milliliters, so let's multiply by 1000, and then after rounding to sig figs we will get 70.1 mL as our answer. (Note: I used the exact values instead of the approximations throughout this explanation, so if you calculate the answer by plugging in these values, it might be slightly off.)

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The graph represents the amount of energy in the

Vitek1552 [10]

Answer: The correct answer is in an exothermic reaction the energy of the product is less than the energy of the reactants.

Explanation: got it

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3 years ago

Nuclear decay is a first-order kinetics process. What is the half-life of a radioactive isotope if it takes 233 minutes for the

-BARSIC- [3]

Answer: The half life of the given radioactive isotope is 43.86 minutes

Explanation:

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken for decay process = 233 minutes

$[A_o]$ = initial amount of the reactant = 0.500 M

[A] = amount left after decay process = 0.0125 M

Putting values in above equation, we get:

$k=\frac{2.303}{233}\log\frac{0.500}{0.0125}\\\\k=0.0158min^{-1}$

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half-life of the reaction = ?

k = rate constant = $0.0158min^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.0158min^{-1}}=43.86min$

Hence, the half life of the given radioactive isotope is 43.86 minutes

8 0

3 years ago

What is responsible for the larger size of an anion in comparison with the atom from which it is form?

GarryVolchara [31]

Answer:

electron-electron repulsion

Explanation:

When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.

In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.

To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.

Download pdf

4 0

3 years ago

Which is the formula for iodine tribromide? IBr IB3 IBr3 I3Br

azamat

Answer: ( C ) IBr3

Explanation: Iodine tribromide formula is IBr3

***If you found my answer helpful, please give me the brainliest, please give a nice rating, and the thanks ( heart icon :) ***

6 0

3 years ago

Read 2 more answers

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago