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Furkat [3]
3 years ago
12

Is SiCI4 a ionic or covalent

Chemistry
2 answers:
Tom [10]3 years ago
7 0

The bonds of the SiCl4 are covalent because Si is not a metal

ololo11 [35]3 years ago
7 0

SiCl4 is a Covalent

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1.) A metal...
Vera_Pavlovna [14]

Answer:

B

Explanation:

Bad conductor of elecricity

5 0
3 years ago
There is water on the pan of the scale as you measure the mass of an object. If you were to ignore the water, what would be the
mihalych1998 [28]
Remember that density refers to the "mass per unit volume" of an object.

So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.

In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is

As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

The effect on density would be that it would erroneously appear to be greater

Hope this helps!

Good luck
6 0
4 years ago
What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =
ValentinkaMS [17]

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Learn more about Concentration here brainly.com/question/14500335

#SPJ4

7 0
2 years ago
Which of the following polyatomic ions will form an ionic compound with two sodium ions? CO32− HCO31− NO21− NO31−
BaLLatris [955]

Answer:

CO32−

Explanation:

We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.

When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;

2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)

8 0
4 years ago
Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,
vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP  weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0
4 years ago
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