All categories

3 years ago

Horsetail are helpful for treating

Chemistry

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

urinary tract infections,

urinary tract infections, loss of bladder control (urinary incontinence),

urinary tract infections, loss of bladder control (urinary incontinence), wounds,

urinary tract infections, loss of bladder control (urinary incontinence), wounds, and many other conditions, but there is no good scientific evidence to support these uses.

You might be interested in

Describe how to prepare 10 ml of 5, 10, 15, and 20 micro M CV solution using a 25 microM CV stock solution

zalisa [80]

A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a stock solution whose concentration is known.

4 0

3 years ago

How many electrons may be accommodated in the first three energy levels?

viktelen [127]

Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. Some points will be nice

3 0

3 years ago

Read 2 more answers

How many valence electrons does rubidiumn have?

vlada-n [284]

Answer:

Explanation:

I believe it really only has 1 valence electrons

7 0

2 years ago

If the fugacity coefficient of the components of a binary mixture are = 0.784, and =0.638 and mole fraction of component 1 is 0.

Reika [66]

Answer : The expression for the fugacity coefficient $\ln \phi$ , for the mixture is, -0.3669.

Explanation : Given,

Fugacity coefficient of component 1 = 0.784

Fugacity coefficient of component 2 = 0.638

Mole fraction of component 1 = 0.4

First we have to calculate the mole fraction of component 2.

As we know that,

$\text{Mole fraction of component 1}+\text{Mole fraction of component 2}=1$

$\text{Mole fraction of component 2}=1-0.4=0.6$

Now we have to calculate the expression for the fugacity coefficient $\ln \phi$ .

Expression used :

$\ln \phi=X_1\ln \phi_1+X_2\ln \phi_2$

where,

$\phi$ = fugacity coefficient

$\phi_1$ = fugacity coefficient of component 1

$\phi_2$ = fugacity coefficient of component 2

$X_1$ = mole fraction of of component 1

$X_2$ = mole fraction of of component 2

Now put all the give values in the above expression, we get:

$\ln \phi=0.4\times \ln(0.784)+0.6\times \ln(0.638)$

$\ln \phi=-0.3669$

Therefore, the expression for the fugacity coefficient $\ln \phi$ , for the mixture is, -0.3669.

7 0

3 years ago

How many moles of Cd and of N are contained in 132.4 g of Cd(N03)2-4H20? (b) How many molecules of water of hydration are in thi

ycow [4]

Answer: The given amount of $Cd(NO_3)_2.4H_2O$ contains 0.430 moles of Cd, 0.860 moles of N and $2.59\times 10^{23}$ molecules of water.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Cd(NO_3)_2.4H_2O$ = 132.4 g

Molar mass of $Cd(NO_3)_2.4H_2O$ = 308.5 g/mol

Plugging values in equation 1:

$\text{Moles of }Cd(NO_3)_2.4H_2O=\frac{132.4g}{308.5g/mol}=0.430 mol$

1 mole of $Cd(NO_3)_2.4H_2O$ contains 1 mole of Cd, 2 moles of nitrogen atom (N), 10 moles of oxygen atom (O) and 8 moles of hydrogen atom (H).

So, 0.430 moles of $Cd(NO_3)_2.4H_2O$ will contain = $(1\times 0.430) = 0.430mol$ of Cd and $(2\times 0.430)=0.86mol$ of N

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, 0.430 moles of $Cd(NO_3)_2.4H_2O$ will contain = $\frac{6.022\times 10^{23}}{1mol}\times 0.430mol=2.59\times 10^{23}$ number of water molecules.

Hence, the given amount of $Cd(NO_3)_2.4H_2O$ contains 0.430 moles of Cd, 0.860 moles of N and $2.59\times 10^{23}$ molecules of water.

4 0

3 years ago