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valentina_108 [34]
3 years ago
7

What is the solution to the problem rounded to the correct number of significant figures? 8.01 × 4.1 = ? A. 32.8 B. 32.84 C. 32.

841 D. 33
Chemistry
2 answers:
Murljashka [212]3 years ago
8 0

Answer:

d

Explanation:

kicyunya [14]3 years ago
7 0

<u>Answer:</u> The correct option is Option D.

<u>Explanation:</u>

Significant figures are defined as the figures that represent the digits of a number which carry an important contribution to the numerical value, starting with the first non-zero digit. For Example: 105.268 has 6 significant figures because every digit carry its own contribution towards the numerical value.

Whenever there is multiplication, the answer will contain the same number of significant figures as there are in the least precise numerical value.

Significant figures in 8.01  are 3 and that in 4.1 are 2.

So, the answer of these two numerical value will contain 2 significant figures.

8.01\times 4.1=32.841\approx 33

Hence, the correct option is Option D.

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Which statement best describes the liquid state of matter?
Andrei [34K]

Answer:

There are no acceptable descriptions at all on that list of choices.

5 0
4 years ago
Read 2 more answers
How many moles of aspartame are present in 4.50 mg of aspartame?
Alekssandra [29.7K]

Answer:- 1.53*10^-^5mol .

Solution:- Aspartame is a artificial sweetener with molecular formula C_1_4H_1_8N_2O_5 and it's molecular mass is  294.3 gram per mol.

We have been given with 4.50 mg of aspartame and asked to calculate the moles. For mass to moles conversion we divide the given mass in grams by molar mass.

Since, the mass is given in mg, we need to convert it to g first.

1 mg = 0.001 g

So, 4.50mg(\frac{0.001g}{1mg})

= 0.00450 g

Now  let's calculate the moles on dividing the grams by molar mass as:

0.00450g(\frac{1mol}{294.3g})

= 1.53*10^-^5mol

Hence, 1.53*10^-^5mol are present in 4.50 mg of aspartame.

4 0
4 years ago
Chlorine has two isotopes, 35Cl and 37Cl that have the respective percent abundance's of 75.77% and 24.23% and masses of 34.9688
AnnyKZ [126]

Answer:

The answer to your question is  35.4527 g

Explanation:

Data

Cl-35    abundance = 75.77 %    mass Cl-35 = 34.9688

Cl-37     abundance = 24.23 %   mass Cl-37 = 36.9659

Process

1.- Convert the percents to decimals

   75.77% = 75.77/100 = 0.7577

   24.23% = 24.23/100 = 0.2423

2.- Find the atomic weight

    Average atomic weight = Abundance 1 x Mass Cl-35 +

                                               Abundance 2 x Mass Cl-37

Substitution

     Average atomic weight = (0.7577)(34.9688) + (0.2423 x 36.9659)

Simplification

     Average atomic weight = 26.4959 + 8.9568

Result

      Average atomic weight = 35.4527 g

3 0
3 years ago
7.
Zielflug [23.3K]

Answer:

D. 18.62

Explanation:

8 0
3 years ago
A large balloon contains 233 L of neon at 22.0C and 760. Torr. If the balloon ascends to an altitude where the temperature is -
shepuryov [24]

Answer: The volume in the balloon at the higher altitude is 260 L

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 760 torr

P_2 = final pressure of gas = 511 torr

V_1 = initial volume of gas = 233 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 22.0^0C=(22.0+273)K=295K

T_2 = final temperature of gas = -52.0^0C=(-52.0+273)K=221K

Now put all the given values in the above equation, we get:

\frac{760\times 233}{295}=\frac{511\times V_2}{221}

V_2=260L

The volume in the balloon at the higher altitude is 260 L

3 0
3 years ago
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