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3 years ago

Please explainnnnnnnnnn

Chemistry

2 answers:

deff fn [24]3 years ago

7 0

Volume=10ml

Density=1.5g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf\longmapsto Mass=Density\times Volume$

$\\ \sf\longmapsto Mass=1.5(10)$

$\\ \sf\longmapsto Mass=15g$

zysi [14]3 years ago

5 0

The answer is 15g
The mass of a substance when given the density and volume can be found by using the formula
Mass = Density x Volume
From the question
Density of chloroform=1.5 g/ml
Volume=10mL
We have
Mass=1.5x10
We have the final answer as
15g

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Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution

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Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

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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

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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.

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4 years ago

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3 years ago

A group 2 metal carbonate has a mass of 84 g/mol. Identify the group 2 metal X using its chemical formula.

zepelin [54]

Answer:

Magnesium

0.003mole

Explanation:

The problem here entails we find the metal in the carbonate.

For group 2 member, let the metal = X;

The carbonate is XCO₃;

If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol

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O = 16g/mol

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Number of moles = $\frac{mass}{molar mass}$

Number of moles = $\frac{0.3}{100}$ = 0.003mole

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3 years ago