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Mnenie [13.5K]
2 years ago
10

Is anyone good at chemistry if so can someone help me please ? (NO LINKS)

Chemistry
1 answer:
V125BC [204]2 years ago
4 0

So, litmus paper is a qualitative tool for assessing the acidity or basicity of a substance (usually a solution). In general, blue litmus turns red in the presence of an acid, and red litmus turns blue in the presence of a base. They can't really tell you much more than that.

The solutions that are most likely acids are those that turn blue litmus red <em>and </em>do not evoke a color change in red litmus. A solution that turns red litmus blue <em>or </em>does not evoke a color change in blue litmus is likely not an acid. Using these criteria, solutions 4 and 7 are most likely acids since they both turn blue litmus red (and they cause no color change in red litmus).

The correct answer choice would thus be D.

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_____ are added to the soil to improve the fertility of the soil​
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Answer:

Fertilisers

And also organic compounds

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2 years ago
Which of the following would not be classified as a mineral?
Setler79 [48]
B. a sugar crystal

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How many grams of sodium are needed to react completely with 19
gtnhenbr [62]

Answer:

46 g of sodium

Explanation:

Sodium reacts vigorously with fluoride gas to form NaF as shown

2Na (s) + F2 (g) ------> 2NaF (s)            (Na = 23, F = 19)

2 moles of Na reacts with 1 mole of F2 to produce NaF

By calculating the molar mass of the elements involved in the reaction then multiplied by the mole, the mass can be obtained.

23 * 2 g/mol of Na reacts with 1 * 19 g/mol of F2

46 g/mol of Na reacts with 19 g/mol of F2 to produce NaF

Since the mole ratio is 2 to 1 and 19 g of F2 is used for the reaction, 46 g of sodium will be consumed for the reaction to be achieved.

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3 years ago
Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure
Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

\Delta T_f=i\times k_f\times m

where,

\DeltaT_b = change in freezing point

k_b = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of KCl will be,

KCl\rightarrow K^++Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl_2(aq) at standard pressure is lower.

4 0
3 years ago
Calculate the value of 0.1394÷0.004 showing all workings​
Lesechka [4]

Answer:

34.85

Explanation:

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3 years ago
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