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3 years ago

He molecular formula mass of this compound is 180 amu . what are the subscripts in the actual molecular formula?

Chemistry

1 answer:

mezya [45]3 years ago

5 0

I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.

Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol

Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6

Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>

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For the hypothetical compound AX3, which statement is true?

Artist 52 [7]

Answer:

b because it says it in the question

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2 years ago

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o

Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

$Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}$

Now, we are able to solve for m:

$m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} } = 9.70*10^4 g Na=97 Kg Na$ =97 000 g Na

7 0

3 years ago

Read 2 more answers

If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30.17}\ year^{-1}$

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Initial concentration $[A_0]$ = ?

Final concentration $[A_t]$ = 12.5 grams

Applying in the above equation, we get that:-

$12.5\ g=[A_0]e^{-0.02297\times 90.6}$

$[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g$

<u>Mass of original sample = 100 g</u>

8 0

3 years ago

Which of the following rules should you follow to balance chemical equations?

Firdavs [7]

Answer:

c. add coefficients as needed

Explanation:

A chemical equation is defined as the equation that shows changes in a chemical reaction. A chemical equation consist of reactant and product, reactant is at left side of the arrow and product is at right side of the arrow.

Reactant => Product

While balancing a chemical equation, the basic rule is to balance the coefficient as required. Coefficient represents the number of molecules and is used at front of a chemical symbol. Change in coefficient helps balance the number of atoms or molecules of the substances on both the sides of the arrow.

Subscripts are never allowed to change because it can change the chemical involved in the reaction.

Hence, the correct answer is "c. add coefficients as needed".

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2 years ago

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2 years ago