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2 years ago

12

. H ow many milliliters of water must be added to 250 mL of a 25% w/v stock solu- tion of sodium chloride to prepare a 0.9% w/v

sodium chloride solution

1 answer:

vaieri [72.5K]2 years ago

4 0

The right answer would be

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HELP HELP HELP PLSSS

andrew11 [14]

Answer:

the answer is c kept in blue and with light

7 0

3 years ago

Calculate the wavelength of a football (425g) thrown by an NFL quarterback traveling at 50mph

AnnyKZ [126]

Wavelength is 6.976 x 10^ -35 m

Explanation:

In this, we can use De Broglie’s equation. This equation is the relationship between De Broglie’s wavelength, velocity and the mass of a moving object. In this equation, we are using plank's constant which is 6.626 x 10^-34 m^2 kg/s.

We know that one mile per hour is equivalent to 0.447 M/S.

And One gram is equivalent to 10^-3 kg.

De Broglie’s wavelength = λ ( wave length) = Plank’s constant/ Mass x velocity

λ ( wave length) = 6.626 x 10^ -34/ (425 x10^-3) x ( 50 x 0.447)

= 6.626 x 10^ -34/ 0. 425 x 22.35

= 6.626 x 10^ -34/ 9.498

= 6.976 x10^ -35 m

So, the wavelength of the football will be 6.976 x 10^ -35 m

7 0

3 years ago

What is oxidation number of S in H2SO5??

mote1985 [20]

★ « <em><u>what is oxidation number of S in H2SO5??</u></em><em><u> </u></em><em><u>»</u></em><em><u> </u></em><em><u>★</u></em>

<em><u>it's </u></em><em><u> </u></em><em><u>6</u></em><em><u>!</u></em><em><u>!</u></em>

Explanation:

<em>Oxidation number of S in H2SO5 is 6 .</em>

3 0

2 years ago

Read 2 more answers

trial 2, 2.68 g/cm/3 trial 3, 2.84g/cm/3. aluminum has a density of 2.70g/cm/3 calcute the percent error for each trial.

serg [7]

Answer:

trial 2: 0.74

trial 3: 5.19

I think but the equation for sloving percent error is:

(true value - determined value)/true value * 100

Explanation:

6 0

3 years ago

6) A volume of 473 mL of oxygen was collected at 27°C. What volume would the oxygen occupy at

sertanlavr [38]

Answer : The volume of oxygen occupy at 173° would be, 703.2 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=473mL\\T_1=27^oC=(27+273)K=300K\\V_2=?\\T_2=173^oC=(173+273)K=446K$

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

$\frac{473mL}{300K}=\frac{V_2}{446K}\\\\V_2=703.2mL$

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL

6 0

2 years ago