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1 year ago

How many moles are 3. 01 x 1022 atoms of magnesium?

Chemistry

1 answer:

hram777 [196]1 year ago

3 0

Answer:

0.05 moles

Explanation:

There are about 6.02 * 10^23 atoms in a mole, so in the given sample, there are (3.01*10^22)/(6.02*10^23)=(1/2)(1/10)=0.05 moles.

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4NH3 + 5O2-> 4NO + 6H2O How many molesof NO are formed if 824g of NH3 react ?

ZanzabumX [31]

824 g NH3 (1 mol/17 g NH3) (4 mol NO/4mol NH3)

48.47 moles NO

4 0

2 years ago

Which of the following is the best thermal conductor? metal wood paper air

Ganezh [65]

Answer:

metal

Explanation:

5 0

2 years ago

Read 2 more answers

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

Answer: The correct answer is Option 5.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

7 0

3 years ago

During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul

gregori [183]

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

7 0

3 years ago

Given the partial equation: NO3− Pb2 → NO2 Pb4 , balance the reaction in acidic solution using the half-reaction method and fill

Fed [463]

Answer : The balanced reaction in acidic solution is,

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

Explanation :

The given partial equation is,

$NO_3^-+Pb^{2+}\rightarrow NO_2+Pb^{4+}$

First we have to separate into half reaction. The two half reactions are:

$NO_3^-\rightarrow NO_2$

$Pb^{2+}\rightarrow Pb^{4+}$

Now we have to balance the half reactions in acidic medium, we get:

$NO_3^-+2H^++1e^-\rightarrow NO_2+H_2O$ ............(1)

$Pb^{2+}\rightarrow Pb^{4+}+2e^-$ ............(2)

Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2, we get

$2NO_3^-+4H^++2e^-\rightarrow 2NO_2+2H_2O$ ...........(3)

Now we have to add both the half reactions (2) and (3), we get the final balanced chemical reaction.

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

3 0

2 years ago