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Arisa [49]
4 years ago
5

Which equations are in standard form? Check all that apply

Chemistry
1 answer:
aliya0001 [1]4 years ago
4 0

Answer:

2x+3y=-6

4x+3y=12

3x-y=5

5x+3y=1

Explanation:

standard form: ax + by = C

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The mass of a nucleus is 0.042 amu less than the sum of the masses of 3 protons and 4 neutrons. The binding energy per nucleon i
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The answer is 10 mev.                                           hope this helps
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3 years ago
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Calculate the volume in ) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO 3 (s) . There is an excess of CaCl 2. What’s the vo
luda_lava [24]

Answer:

100 mL of Na2CO3

Explanation:

We'll begin by calculating the number of mole in 1 g of CaCO3. This can be obtained as follow:

Mass of CaCO3 = 1 g

Molar mass of CaCO3 = 100.09 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 1/100.09

Mole of CaCO3 = 0.01 mole

Next, we shall determine the number of mole of Na2CO3 needed to produce 0.01 mole of CaCO3.

This is illustrated below:

Na2CO3 + CaCl2 —> 2NaCl + CaCO3

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CaCO3.

Therefore, 0.01 mole of Na2CO3 will also react to produce 0.01 mole of CaCO3.

Next, we shall determine the volume of Na2CO3 needed for the reaction as illustrated below:

Mole of Na2CO3 = 0.01 mole

Molarity of Na2CO3 = 0.1 M

Volume of Na2CO3 solution needed =?

Molarity = mole /Volume

0.1 = 0.01 / volume of Na2CO3

Cross multiply

0.1 × volume of Na2CO3 = 0.01

Divide both side by 0.1

Volume of Na2CO3 = 0.01 / 0.1

Volume of Na2CO3 = 0.1 L

Finally, we shall convert 0.1 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.1 L = 0.1 L × 1000 mL / 1 L

0.1 L = 100 mL

Thus, 0.1 L is equivalent to 100 mL.

Therefore, 100 mL of Na2CO3 is needed for the reaction.

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greenhouse effects are molecules that trap heat on earth atmosphere

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Why are 35 17 cl bad 37 17 cl isotopes of the element of chlorine? ​
Lisa [10]

Answer:

Chlorine-35 and chlorine-37 are not different elements because an atom of chlorine-35 and an atom of chlorine-37 each contain the same number of protons. The number of protons an atom has, also known as the atom's atomic number, determines which element it is. All atoms which contain 17 protons are called chlorine atoms. Adding or removing a proton from an atom's nucleus changes that atom's atomic number and creates a different element.

Chlorine-35 and chlorine-37 are both isotopes of the element chlorine. The number after the name 'chlorine' is called the mass number. The mass number is a tally of the number of protons and the number of neutrons in an atom's nucleus. Since all atoms of chlorine contain 17 protons, chlorine-35 and chlorine-37 differ in the number of neutrons each one has. An atom of chlorine-35 contains 18 neutrons (17 protons + 18 neutrons = 35 particles in the nucleus) while an atom of chlorine-37 contains 20 neutrons (17 protons + 20 neutrons = 37 particles in the nucleus).

Adding or removing a neutron from an atom's nucleus creates isotopes of a particular element. Why does changing the number of protons in an atom change which element that atom is but changing the number of neutrons doesn't? Protons carry a positive charge. Each proton in an atom's nucleus must be balanced with a negatively charged electron in one of the 'shells' outside the nucleus. The number of electrons in an atom's outer shell determines the atom's chemical properties. Adding or removing protons changes the number and arrangement of electrons in the outer shell which changes how that atom reacts with other atoms. Neutrons don't change things greatly because they do not carry an electrical charge. Neutrons can be added or removed from an atom and the electrons around the atom really don't care that much.

3 0
3 years ago
When solid calcium carbonate is heated, it decomposes according to the reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) Kp = 0.50 A sample o
blagie [28]

Explanation:

Relation between K_{p} and K_{c} is as follows.

            K_{p} = K_{c} [RT]^{\Delta n}

Given,   temperature = 830^{o}C = (830 + 273) K = 1103 K

            R = 8.314 J/mol K

       \Delta n = 1 - 0 = 1

Now, putting the given values into the above formula as follows.

              K_{p} = K_{c} [RT]^{\Delta n}

                 0.5 = K_{c} \times (8.314 \times 1103)^{1}

                     K_{c} = 5.452 \times 10^{-3}

ICE table for the given reaction will be as follows.

                  CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)

Initial:           c                        -          -  

Equilibrium:  (c - x)               x         x

       K_{c} = [CO_{2}]

          5.452 \times 10^{-3} = x

Hence, same amount of CaO is produced.

Moles of CaO = 5.452 \times 10^{-3}

    Mass of CaO = 5.452 \times 10^{-3} \times 57 g/mol

                          = 0.310 g

Thus, we can conclude that 0.310 g of CaO produced when equilibrium is established.

8 0
3 years ago
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