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Leya [2.2K]
3 years ago
7

How many sigfigs in 10.0 ? Your answer I

Chemistry
1 answer:
alina1380 [7]3 years ago
7 0

Answer:

3

Explanation:

10.0 has 3 significant figures

You might be interested in
How many electrons does an ion of magnesium have if it’s represented by the symbol shown below? 25Mg2+
Galina-37 [17]

Answer:

10 electrons.

Explanation:

Hello!

In this case, since a normal atom of magnesium has 12 protons and electrons when is not in the form of a cation, since it is a metal that tends to lose electrons when ionized, and therefore it becomes positively charged, we infer  that in form of:

Mg^{2+}

It only has 10 electrons.

Best regards!

7 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
4 years ago
Determine the number of sub shells and orbital in the fifth main shell​
Black_prince [1.1K]

Answer: maximum no. is 8 in 5th shell

Explanation:

6 0
3 years ago
A 45.9 g sample of a metal is heated to 95.2°c and then placed in a calorimeter containing 120.0 g of water (c = 4.18 j/g°c) at
VladimirAG [237]

Answer:

Iron

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the metal;

= 45.9 g × c ×(95.2 -24.5) , where c is the specific heat capacity of the metal

= 3245.13c joules

Heat absorbed by water;

= 120 g × 4.18 J/g°C × (24.5-21.6)

= 1454.64 joules

Therefore;

3245.13c joules = 1454.64 joules

c = 1454.64/3245.13

  = 0.448 J/g°C

The specific heat capacity of the  metal sample is 0.448 J/g°C. The metal use is most likely, Iron.

8 0
3 years ago
BRAINLIEST TO CORRECT only answer number 10 please
IRINA_888 [86]

Answer:

they are softer as you go deeper

Explanation:

because the core is liquid

5 0
2 years ago
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