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Bond [772]
3 years ago
14

How many grams of carbonic acid were produced by the 3.00g sample of NaHCO3

Chemistry
1 answer:
Alexxx [7]3 years ago
8 0
I dont know what subject is this
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If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

Learn more about molarity: brainly.com/question/10053901

6 0
1 year ago
What experimental evidence can you provide that the product isolated is 1-bromobutane?
Blizzard [7]
The experimental evidence that you could provide that <span> the product isolated is 1-bromobutane would be the changes that happens that are observable by the naked eye. Hope this helps. Have a nice day. Feel free to ask more questions.</span>
5 0
3 years ago
Read 2 more answers
In which part of a chemical reaction do bonds break in which part do they form?
cricket20 [7]
<span>Chemical bonds are broken down in the reactants while a reaction is occurring. Products are the new chemical bonds that are formed.</span>
3 0
3 years ago
Which of these is an isoelectronic series? question 7 options: 1) na+, k+, rb+, cs+ 2) k+, ca2+, ar, s2– 3) na+, mg2+, s2–, cl–
-BARSIC- [3]
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:

Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6, Ar = 18

A. Na⁺: 11-1 = 10 electrons
     K⁺: 19 - 1 = 18 electrons
     Rb⁺: 37-1 = 36 electrons

B. K⁺: 19 - 1 = 18 electrons
    Ca²⁺: 20 - 2 = 18 electrons
    Ar:  18 electrons
    S²⁻:  16 +2 = 18 electrons

C. Na⁺: 11-1 = 10 electrons
    Mg²⁺: 12 - 2 = 10 electrons
     S²⁻:  16 +2 = 18 electrons

D. Li=3 electrons
    Be=4 electrons
    B=5 electrons
    C=6 electrons

The answer is letter B.
6 0
3 years ago
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0
2 years ago
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