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2 years ago

This is the correct category, right?

Chemistry

1 answer:

Soloha48 [4]2 years ago

5 0

Answer:

32.84002

Explanation:

average atomic mass= F1A1 + F2A2 + F3A3 so that means 60/100*34.456+23/100*37.457+9/100*39.459

You might be interested in

What is the mass of 3.01 x 10^21 atoms of Co

OlgaM077 [116]

Answer:

0.295 g Co

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

3.01 × 10²¹ atoms Co

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Co - 58.93 g/mol

Step 3: Convert

$3.01 \cdot 10^{21} \ atoms \ Co(\frac{1 \ mol \ Co}{6.022 \cdot 10^{23} \ atoms \ Co} )(\frac{58.93 \ g \ Co}{1 \ mol \ Co} )$ = 0.294552 g Co

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.294552 g Co ≈ 0.295 g Co

5 0

3 years ago

Arrange the following aqueous solutions in order of decreasing freezing point: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl2, 0.15 m

GarryVolchara [31]

Answer: 0.35 m $NaCl$ > 0.20 m $MgCl_2$ >0.10 m $Na3PO_4$ > 0.15 m $CH_3COOH$ > 0.15 m $C_6H_{12}O_6$

Explanation:

Depression in freezing point:

$T_f^0-T_f=i\times k_b\times m$

where,

$T_f$ = freezing point of solution

$T^o_f$ = freezing point of solvent

$k_f$ = freezing point constant

m = molality

1. For 0.10 m $Na3PO_4$

$Na_3PO_4\rightarrow 3Na^++PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.10=0.40$

2. For 0.35 m $NaCl$

$NaCl\rightarrow Na^++Cl^-$

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be $2\times 0.35=0.70$

3. For 0.20 m $MgCl_2$

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.20=0.60$

4. For 0.15 m $C_6H_{12}O_6$

, i= 1 as it is a non electrolyte and does not dissociate to give ions.

5. For 0.15 m $CH_3COOH$

$CH_3COOH\rightarrow CH_3COO^-+H^+$

, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have $2\times 0.15=0.30$

As concentration is highest for 0.35 m $NaCl$ , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m $C_6H_{12}O_6$ , freezing point depression will be lowest and thus has highest freezing point

8 0

3 years ago

What is newtons 3rd law

coldgirl [10]

Newton’s third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction. Hope this helps.

3 0

3 years ago

Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10

Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

(1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

$p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL$

(2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

$p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3} \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL$

(3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error= $\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100$

%error= $\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100$

%error=2.44 %

7 0

3 years ago

A group of students is investigating how the addition of salt impacts the floatation of an egg in water. Four identical cups wer

soldier1979 [14.2K]

Answer: The correct answer is Option 1.

Explanation:

Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.

We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.

Hence, the correct answer is Option 1.

4 0

3 years ago