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nadya68 [22]
1 year ago
6

Which of the following occurs when sodium and chlorine bond together to form sodium

Chemistry
1 answer:
Mumz [18]1 year ago
7 0
(a) chlorine gains an electron from a sodium atom
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Suppose that you are performing a titration on a monoprotic acid. Titration of this monoprotic acid required a volume of 0.1L of
Alex73 [517]

Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions

Explanation:

The monoprotic acid (HA) has a valency of 1 and diprotic acid  (H_2A) has a valency of 2.

As the concentration and volume of the diprotic acid and the monoprotic acids are equal.

The neutralization reaction for monoprotic acid is:

HA+BOH\rightarrow BA+H_2O

The neutralization reaction for diprotic acid is:

H_2A+2BOH\rightarrow B_2A+2H_2O

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.

6 0
3 years ago
What metal will have properties most similar to those of chromium (Cr)? Why?
garri49 [273]
That element is manganese. As they are in same horizontal row (period) and are next to each other. That is why they show same properties. 

Hope this helps xox :)
5 0
3 years ago
To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
poizon [28]
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


7 0
3 years ago
How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
salantis [7]

Answer:

85.34g of NH3

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

Step 3:

Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

3 0
2 years ago
The student adds 0.0010 mol of NaOH(s) to solution Y, and adds 0.0010 mol of NaOH(s) to solution Z. Assume that the volume of ea
julia-pushkina [17]

Answer:

The answer is in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Buffers are able to avoid the pH change of a solution when strong acid or bases are added (As NaOH).

Based on the experiment, it is possible that the solution Z was a buffer and Y another kind of solution. For this reson, pH of the solution Y changes much more than the pH of solution Z changes despite the amount of NaOH added is the same in both solutions.

6 0
2 years ago
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