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satela [25.4K]
4 years ago
6

Dextrose 25% 1000 ml was ordered, you have only dextrose 70% solution available. how much of the dextrose 70% solution and steri

le water will you use to fill this order?
Chemistry
1 answer:
Varvara68 [4.7K]4 years ago
4 0
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 70%
c2 = 25%
V1 = ?
V2 = 1000 mL
When we plug values into the equation, we get following:
70 x V1 = 25 x 1000
V1 = 333.33 mL
V(H2O)=V2-V1= 1000-333.33=666.67 mL

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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

5 0
4 years ago
Read the scientific question below.
svetlana [45]
If we fertilize a plant, then its height increases fast. Always use if then format
8 0
3 years ago
How is solar power used to generate electricity?
musickatia [10]
This is the shortest answer, you can google: net meter, inverter, solar panels and the roof system for a shorter one.

The roof system

In most solar systems, solar panels are placed on the roof. An ideal site will have no shade on the panels, especially during the prime sunlight hours of 9 a.m. to 3 p.m.; a south-facing installation will usually provide the optimum potential for your system, but other orientations may provide sufficient production. Trees or other factors that cause shading during the day will cause significant decreases to power production. The importance of shading and efficiency cannot be overstated. In a solar panel, if even just one of its 36 cells is shaded, power production will be reduced by more than half. Experienced installation contractors such as NW Wind & Solar use a device called a Solar Pathfinder to carefully identify potential areas of shading prior to installation.

Not every roof has the correct orientation or angle of inclination to take advantage of the sun's energy. Some systems are designed with pivoting panels that track the sun in its journey across the sky. Non-tracking PV systems should be inclined at an angle equal to the site’s latitude to absorb the maximum amount of energy year-round. Alternate orientations and/or inclinations may be used to optimize energy production for particular times of day or for specific seasons of the year.

Solar panels

Solar panels, also known as modules, contain photovoltaic cells made from silicon that transform incoming sunlight into electricity rather than heat. (”Photovoltaic” means electricity from light — photo = light, voltaic = electricity.)

Solar photovoltaic cells consist of a positive and a negative film of silicon placed under a thin slice of glass. As the photons of the sunlight beat down upon these cells, they knock the electrons off the silicon. The negatively-charged free electrons are preferentially attracted to one side of the silicon cell, which creates an electric voltage that can be collected and channeled. This current is gathered by wiring the individual solar panels together in series to form a solar photovoltaic array. Depending on the size of the installation, multiple strings of solar photovoltaic array cables terminate in one electrical box, called a fused array combiner. Contained within the combiner box are fuses designed to protect the individual module cables, as well as the connections that deliver power to the inverter. The electricity produced at this stage is DC (direct current) and must be converted to AC (alternating current) suitable for use in your home or business.

Inverter

The inverter is typically located in an accessible location, as close as practical to the modules. In a residential application, the inverter is often mounted to the exterior sidewall of the home near the electrical main or sub panels. Since inverters make a slight noise, this should be taken into consideration when selecting the location.

The inverter turns the DC electricity generated by the solar panels into 120-volt AC that can be put to immediate use by connecting the inverter directly to a dedicated circuit breaker in the electrical panel.

The inverter, electricity production meter, and electricity net meter are connected so that power produced by your solar electric system will first be consumed by the electrical loads currently in operation. The balance of power produced by your solar electric system passes through your electrical panel and out onto the electric grid. Whenever you are producing more electricity from your solar electric system than you are immediately consuming, your electric utility meter will turn backwards!

Net meter

In a solar electric system that is also tied to the utility grid, the DC power from the solar array is converted into 120/240 volt AC power and fed directly into the utility power distribution system of the building. The power is “net metered,” which means it reduces demand for power from the utility when the solar array is generating electricity – thus lowering the utility bill. These grid-tied systems automatically shut off if utility power goes offline, protecting workers from power being back fed into the grid during an outage. These types of solar-powered electric systems are known as “on grid” or “battery-less” and make up approximately 98% of the solar power systems being installed today.

5 0
3 years ago
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
4 years ago
For a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k, calculate the inner diameter if you are designing for
bonufazy [111]

The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m

<h3>What is Stack Height ?</h3>

Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.

If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.

Given is a steel stack that exhausts 1,200 cu.m/min of gases

P= 1 atm and

T= 400 K

maximum expected wind speed at stack height of 12 m/s

The formula for the diameter of chimney

\rm d=\sqrt{\dfrac{4Q}{\pi v} }

Q =1200 cu.m/min

= 1200 * 0.0166 = 19.92 cu.m/sec

Velocity = 12m/s

\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\

d= 1.45 m

Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.

To know more about Stack Height

brainly.com/question/24625453

#SPJ4

8 0
3 years ago
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