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3 years ago

_________solution contains sodium hydroxide and copper sulphate

Chemistry

2 answers:

Archy [21]3 years ago

8 0

Biuret reagent. This is the answer

Hunter-Best [27]3 years ago

8 0

Answer: Biuret Reagent

Explanation:

The reagent used in the Biuret Test is a solution of copper sulfate (CuSO4) and sodium hydroxide (NaOH).

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A large highway barrier is 1 meter wide, by 1 meter tall, by 2 meters long.

Romashka [77]

Answer:

when mass is 1×10⁴ Kg then density is 5 g/cm³.

when mass is 104 Kg then density is 5.2 × 10⁻² g/ cm³.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

mass = 1×10⁴ Kg

volume= w ×l× h = 1×2× 1 = 2 m³

density = ?

first of all we will convert the given volume meter cube to cm³:

we know that

2×1000000 = 2 × 10⁶ cm³

Now we will convert the mass into gram.

1 Kg = 1000 g

1×10⁴ × 1000 = 1 ×10⁷ g

Now we will put the values in the formula,

d = m/v

d = 1 ×10⁷ g / 2×10⁶ cm³

d = 0.5 × 10¹ g/cm³

d = 5 g/cm³

If mas is 104 Kg:

104 × 1000 = 104000 g

d= m/v

d = 104000 g / 2×10⁶ cm³

d= 52000 ×10⁻⁶ g/ cm³

d= 5.2 × 10⁻² g/ cm³

7 0

3 years ago

If the internal energy of a system increases but there is no change in temperature, then the system's energy is increasing.

777dan777 [17]

Answer:

kintic

Explanation:

4 0

4 years ago

Read 2 more answers

3 organisms that help prevent infections

love history [14]

Infection control is the discipline concerned with preventing nosocomial or healthcare-associated infection, a practical (rather than academic) sub-discipline of epidemiology. It is an essential, though often underrecognized and undersupported, part of the infrastructure of health care. Infection control and hospital epidemiology are akin to public health practice, practiced within the confines of a particular health-care delivery system rather than directed at society as a whole. Anti-infective agents include antibiotics, antibacterials, antifungals, antivirals and antiprotozoals.[1]

Infection control addresses factors related to the spread of infections within the healthcare setting (whether patient-to-patient, from patients to staff and from staff to patients, or among-staff), including prevention (via hand hygiene/hand washing, cleaning/disinfection/sterilization, vaccination, surveillance), monitoring/investigation of demonstrated or suspected spread of infection within a particular health-care setting (surveillance and outbreak investigation), and management (interruption of outbreaks). It is on this basis that the common title being adopted within health care is "infection prevention and control." (got from google

8 0

3 years ago

I am a liquid metal with 80 electrons

Hitman42 [59]

Answer:

Mercury

Explanation:

Mercury is a liquid element that contains 80 electrons when not charged.

5 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

4 years ago

_________solution contains sodium hydroxide and copper sulphate​

_________solution contains sodium hydroxide and copper sulphate