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3 years ago

PLEASE HELP NOW!!!!!!!

1 answer:

7 0

The number of moles hydrogen produced when 6 moles of sodium are used is calculated as below

2Na + 2H2O = 2NaOH +H2

by use of mole ratio between Na to H2 which is 2:1 the moles of H2 is therefore

= 6 x1/2 = 3 moles of H2

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A solution (HELP ASSAAAAP ROAD TO GRADUATION)

creativ13 [48]

Hey there! A simple explanation is below.

Answer:

D) is a single phase homogeneous mixture.

Explanation:

A solution is a form of homogenous combination made up of two or more components in chemistry. A solute is a material that is dissolved in another material, known as a solvent, in such a combination. The mixing of a solution takes place at a scale where the effects of chemical polarity are present, resulting in solvation-specific interactions. In most cases, the solution is in the condition of the solvent, because it is most common in the mixture.

3 0

3 years ago

can someone solve me some papers I don't have time and I need to submit today please I will make you brilliant

dem82 [27]

Answer:

submit the papers onto this platform and we will try to help out.

6 0

3 years ago

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 0

3 years ago

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are

vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of $OH^-$ ions is greater. The remaining concentration of $[OH^-]$ ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

$[HNO_3]$ = 0.200 M

$[Ca(OH)_2]$ =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of $H^+$ ion from both the acids

moles of $H^+$ in HCl

$HCl\rightarrow {H^+}+Cl^-$

if 1 L of $HCl$ solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 $\times$ 0.1 moles= 0.005 moles (1L=1000mL)

moles of $H^+$ in HCl = 0.005 moles

Similarliy

moles of $H^+$ in $HNO_3$

$HNO_3\rightarrow H^++NO_3^-}$

If 1L of $HNO_3$ solution= 0.200 moles

Then 0.1L of $HNO_3$ solution= 0.1 $\times$ 0.200 moles= 0.02 moles

moles of $H^+$ in $HNO_3$ =0.02 moles

so, Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles .....(1)

Step 2: Calculating the total moles of $[OH^-]$ ion from both the bases

Moles of $OH^-\text{ in }Ca(OH)_2$

$Ca(OH)_2\rightarrow Ca^2{+}+2OH^-$

1 L of $Ca(OH)_2$ = 0.0100 moles

Then in 0.5 L $Ca(OH)_2$ solution = 0.5 $\times$ 0.0100 moles = 0.005 moles

$Ca(OH)_2$ produces two moles of $OH^-$ ions

moles of $OH^-$ = 0.005 $\times$ 2= 0.01 moles

Moles of $OH^-$ in $RbOH$

$RbOH\rightarrow Rb^++OH^-$

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 $\times$ 0.100 moles = 0.02 moles

Moles of $OH^-$ = 0.02 moles

so,Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles ....(2)

Step 3: Comparing the moles of both $H^+\text{ and }OH^-$ ions

One mole of $H^+$ ions will combine with one mole of $OH^-$ ions, so

Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles....(1)

Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of $H^+$ ions = total moles of $OH^-$ ions

0.025 moles $H^+$ will neutralize the 0.025 moles of $OH^-$

Moles of $OH^-$ ions is in excess (from 1 and 2)

The remaining moles of $OH^-$ will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of $OH^-$ ions = $\frac{\text{Moles remaining}}{\text{Total volume}}$

$[OH^-]=\frac{0.005}{0.85}=0.0058M$

6 0

4 years ago

For IR radiation with û = 1,130 cm 1, v=__THz

Dmitry [639]

<u>Answer:</u> The frequency of the radiation is 33.9 THz

<u>Explanation:</u>

We are given:

Wave number of the radiation, $\bar{\nu}=1130cm^{-1}$

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

$\bar{\nu}=\frac{1}{\lambda}$

where,

$\bar{\nu}$ = wave number = $1130cm^{-1}$

$\lambda$ = wavelength of the radiation = ?

Putting values in above equation, we get:

$1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm$

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So, $8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m$

The relation between frequency and wavelength is given as:

$\nu=\frac{c}{\lambda}$

where,

c = the speed of light = $3\times 10^8m/s$

$\nu$ = frequency of the radiation = ?

Putting values in above equation, we get:

$\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}$

$\nu=0.339\times 10^{14}Hz$

Converting this into tera Hertz, we use the conversion factor:

$1THz=1\times 10^{12}Hz$

So, $0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz$

Hence, the frequency of the radiation is 33.9 THz

7 0

3 years ago