All categories

3 years ago

How many significant digits are there in the measurement 50,600 mg A)4 B)5 C)6 D)7

Chemistry

2 answers:

Arte-miy333 [17]3 years ago

7 0

Five. The measurement 50,600 mg has five significant digits.

I presume that you are using the comma as a decimal separator.

The rules for significant figures are
1. Nonzero digits are always significant.
2. Any zeros between two significant digits are significant.
3. Final or trailing zeros are significant only if they are to the right of a decimal point.

• According to Rule 1, the 5 and 6 are significant.
• According to Rule 2, the 0 between the 5 and 6 is significant
• According to Rule 3, the final two zeros are significant.

Thus, there are five significant digits in the measurement 50,600 mg.

Note: If the comma is a thousands separator, the number has only three significant digits.

stellarik [79]3 years ago

3 0

Answer:

Explanation:

Significant figures may be defined as the numbers of the digit that can carry to contribute the meaning of the exact measurement resolution. Significant figures follows some specific scientific rules.

All non zero digits are significant figures. The zeros that are present between the non zero digits are significant figures. The zeros that are present after the significant figures are considered as significant. Here, 5 significant figures are 5,0,6,0 and 0. Hence, the total significant figures are 5.

Thus, the correct answer is option (B).

You might be interested in

Of the rock's original uranium-235 remains. how old is the rock

Gemiola [76]

2.1 billion years
2,139,000,000 to be exact

5 0

3 years ago

Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?

kifflom [539]

The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0

3 years ago

Read 2 more answers

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

Answer: The molality of the solution is 0.11 m

Explanation:

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

3 years ago

The average person in the United States is exposed to the following amount of radiations annually. Rank the following source of

Papessa [141]

#1 is air radon, #2 is x-ray, #3 is ground, #4 is cosmic radiation, #5 is TV tube, #6 is weapons test fallout . That's all I got hope I helped!

8 0

3 years ago

Read 2 more answers