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3 years ago

1. Isotopes of are atoms of the same element that vary in

Chemistry

1 answer:

Phantasy [73]3 years ago

6 0

Answer:

1. D

2. B

Explanation:

1. Isotopes vary in neutrons. Due to this variance, the difference in neutrons change the mass of the element.

Note: the amount of protons in an element never changes. The number of protons (the atomic number on the periodic table) identifies the element in question. If you add or subtract protons, you change the element.

2. B-10 and B-11 are two isotopes of the element Boron: One with 10 neutrons, and another with 11 neutrons. Because they contain a variance of neutrons, they do not have the same mass. This cancels choices A and D.

The protons of an element are equal to the atomic number of that element. Borons atomic number is 5 (according to the periodic table), so it contains 5 protons. This cancels choice C as the element with 6 protons is Carbon.

Note: The amount of protons in an atom is balanced out by the same amount of electrons to allow for a neutral charge of the element (i.e., the protons, which contain a positive charge, and electrons, which contain a negative charge, cancel each other out. In the case of Boron: +5 protons - 5 electrons = 0 charge). This is why choice B is correct, as the consideration of an ion is not necessary for this question.

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Changes between a liquid and gas exsample

exis [7]

Here are some examples:-

Water → water vapour

bromine → bromine gas

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2 years ago

Use the chemical equation below to determine how many moles of ammonia

anzhelika [568]

Answer:

If we assume that there will be enough Hydrogen for the reaction to occur, then there will be 8 moles of NH

Explanation:

The balanced equation will look like this:

4N2 + 4H2 -> 8NH

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3 years ago

Are Zombies Living or Non-living Reasoning and Evidence?

Thepotemich [5.8K]

Answer:

I believe that zombies are not living

Explanation:

because zombies don't grow and develop because they are practically mindless. Also, zombies definitely respond to the environment (like when they see a human) but they don't adapt to it

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3 years ago

Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is the number of moles of solute in 1 L of solution; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0

3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

3 years ago