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2 years ago

What could have caused the mass to be higher than the initial mass

Chemistry

2 answers:

ivann1987 [24]2 years ago

6 0

Answer:

Possibly the addition of oxygen from the atmosphere

Explanation:

Amanda [17]2 years ago

6 0

The oxygen comes from the atmosphere

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An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog

dlinn [17]

Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

Platinum (Pt) = 117.4 g

Carbon (C) = 28.91 g

Nitrogen (N) = 33.71 g

Divide by their molar mass

Pt = 117.4 / 195 = 0.602

C = 28.91 / 12 = 2.409

N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

C = 2.409 / 0.602 = 4

N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

8 0

2 years ago

What is the frequency of an electromagnetic wave that has a wavelength of

Korolek [52]

➡ ANSWER

☑ C. 3.5 105 Hz

3 0

2 years ago

A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

Marrrta [24]

Answer:

$V_2=12.1L$

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

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

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

$T_1=20+273=293K\\\\T_2=-40+273=233K$

Then, we obtain:

$V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L$

Best regards!

5 0

2 years ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

3 years ago

The sugar ribose is an important component of RNA. Ribose is 40.0% C, 6.7% H, and 53.3\% by mass. How many grams of each element

Brums [2.3K]

Answer:

C_5H_10O_5

Explanation:

5 0

3 years ago

What could have caused the mass to be higher than the initial mass​

What could have caused the mass to be higher than the initial mass