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3 years ago

Help please the second question. I really need help guys

Chemistry

1 answer:

statuscvo [17]3 years ago

6 0

Answer:

106g/mol

Explanation:

Mass of Na=2×23=46

Mass of C=12

Mass of O=3×16=48

Total mass of Na2CO3=106g/mol

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Calculate the molar mass of ammonium chloride

il63 [147K]

Answer:

53.491 g/mol

Explanation:

Create the chemical compound and find each individual element's molar mass. Lastly, add them up.

3 0

3 years ago

How many molecules are in 68.0 g of H2S

Kryger [21]

The correct answer is 12.044 × 10²³ molecules.

The molecular mass of H₂S is 34 gram per mole.

Number of moles is determined by using the formula,

Number of moles = mass/molecular mass

Given mass is 68 grams, so no of moles will be,

68/34 = 2 moles

1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2

= 12.044 × 10²³ molecules.

3 0

3 years ago

when objects like horseshoe are left outside over time, the iron in them will react with oxygen to produce rust. Is this a chemi

xxTIMURxx [149]

Feros is a chemical property

5 0

3 years ago

Give two test substances that would react with each other to produce salt and water

Mademuasel [1]

Answer:

NaOH +HCl==>Nacl+H2O

KOH+HCl==>KOH+H2O

6 0

3 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

6 0

3 years ago