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DIA [1.3K]
1 year ago
10

An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

off, the pH is adjusted to about 1 and H2S is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and H2S is again added; no precipitate forms. No precipitate forms upon addition of (NH4)2HPO4. The remaining solution shows a yellow color in a flame test.
Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent.
CdS, Pb(NO3)2, HgO, ZnSO4, Cd(NO3)2, and Na2SO4
Chemistry
1 answer:
Karo-lina-s [1.5K]1 year ago
5 0

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

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in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally
Gwar [14]

Answer:

The average atomic mass of bromine is 79.9 amu.

Explanation:

Given data:

Percentage of Br⁷⁹ = 55%

Percentage of Br⁸¹ = 45%

Average atomic mass of bromine = ?

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [55 × 79] + [81 ×45] / 100

Average atomic mass = 4345 + 3645 / 100

Average atomic mass = 7990 / 100

Average atomic mass = 79.9 amu

The average atomic mass of bromine is 79.9 amu.

3 0
2 years ago
Before raw data can be used as scientific evidence, it must be analyzed and summarized. One way of summarizing data is to find t
Nikitich [7]
The mean is simply the arithmetic average of all your raw data. This can be solved methodically by summing up all of the raw data points that you have. Take note how many raw data points you used, because this will be used to divide the sum. You will obtain the mean.
6 0
2 years ago
Anna lives in a city that experiences high precipitation, with an average annual rainfall of 524 millimeters. It is warm all yea
gavmur [86]

Anna lives in a city that is part of the tropical climate types. It has a constantly warm weather, and thus higher humidity, and according to the annual rainfall, it is most probably a rainfall that appears seasonally, not throughout the whole year.

Tim, on the other hand, lives in a city that is part of the dry climate types. It is most probably a place that is deep into the mainland, like the cold deserts of Central Asia, where the temperatures in the summer are high, and in winter are very low. Because of the distance from the sea, the rainfall doesn't reach this places, so they are very dry, and only have symbolic amount of annual rainfall.

6 0
2 years ago
A 7.50 liter sealed jar at 18 °c contains 0.125 moles of oxygen and 0.125 moles of nitrogen gas. what is the pressure in the con
devlian [24]
The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles 
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
4 0
2 years ago
Read 2 more answers
3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con
KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}

c) Half life for first-order kinetics is computed by:

t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}}  =53.3min

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M

Best regards.

6 0
2 years ago
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