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Ymorist [56]
3 years ago
12

Pls help!!plz answer correctly!will give the brainliest!Urgent!!​

Chemistry
2 answers:
blagie [28]3 years ago
7 0

a)

A: Copper

B: CuO

C: \mathrm{CuSO_4}

D: $\mathrm{CuCO_3}$

E: $\mathrm{CO_2}$

F: $\mathrm{Cu(NO_3)_2}$

b)

$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$

c)

$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$

daser333 [38]3 years ago
5 0
Yes listen to the perosn on top yes yes
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What is the formula for hydrophosphoric acid?
34kurt

Answer:

H3PO4 is the formula for hydrophosphoric acid

4 0
3 years ago
What are the charges of families 1, 2, & 13 (boron family) if/when they become Ions
Natalka [10]

Answer:

Explanation:

Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.

Charges on group one.

Hydrogen = +1

Lithium = +1

Sodium = +1

Potassium = +1

Rubidium = +1

Cesium = +1

Francium = +1

Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.

Charges on group two.

Beryllium = +2

Magnesium = +2

Calcium = +2

Strontium = +2

Barium= +2

Radium = +2

Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.

Charges on group 13.

Boron = +3

Aluminium = +3

Gallium = +3

Indium = +3

Thallium= +3

Group 13 elements are also shows +1 charge by losing one valance electron.

8 0
3 years ago
Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0
3 years ago
What are two processes that result in rocks being broken down into smaller pieces !?
Snowcat [4.5K]
<span>"Chemical weathering and physical weathering" would be the correct answer

Chemical weathering breaks down the bonds holding the rocks together, and the physical weathering will crush and break them apart.</span>
7 0
3 years ago
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
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