Which of the following inequalities represents the statement of: a number x increased by 4 is greater than or equal to 19

PLEASE HELP!!!

Aliun [14]

Answer:

a. 8.75 M NaOH

b. 0.425 M CuCl₂ or 0.43 M CuCl₂

c. 0.067 M CaCO₃ or 0.07 M CaCO₃

Step-by-step explanation:

Molality is computed using the formula:

$M = \dfrac{Moles\:of\:solute}{Liters\:of\:solution}$

So first thing you need to do is determine how many moles of solute there are and divide it by the solution in liters.

Converting mass to moles, you need to get the mass of each solute per mole. You can use the periodic table to get the atomic mass (which is the grams per mole of each atom) of each of the elements involved. Then add them up and you will have how many grams per mole of each compound.

1. 35.0g of NaOH in 100ml H₂O

Element number of atoms atomic mass TOTAL

Na 1 x 22.99g/mol = 22.99g/mol

O 1 x 16.00g/mol = 16.00g/mol

H 1 x 1.01g/mol = 1.01g/mol

40.00g/mol

This means that the molecular mass of NaOH is 40.00 g/mol

Then we use this to convert 35.0g of NaOH to moles:

$35.0g \:of\:NaOH \times \dfrac{1\:mole\:of\:NaOH}{40.00g\:of\:NaOH} = \dfrac{35.0\:moles\:of\:NaOH}{40.00}=0.875\:moles\:of\:NaOH$

Now that you have the number of moles we divide it by the solution in liters. Before we can do that you have to conver 100ml to L.

$100ml\times\dfrac{1L}{1000ml} = 0.1 L$

Then we divide it:

$\dfrac{0.875\:moles\:of\:NaOH}{0.1L of solution} = 8.75M\: NaOH$

2. 20.0g CuCl₂ in 350ml H₂O

Element number of atoms atomic mass TOTAL

Cu 1 x 63.55g/mol = 63.55g/mol

Cl 2 x 34.45g/mol = 70.90g/mol

134.45g/mol

$20.g\:of\:CuCl_2\times\dfrac{1\:mole\:of\:CuCl_2}{134.45\:g\:of\:CuCl_2}=0.1488\:moles\:of\:CuCl_2$

350ml = 0.350L

$\dfrac{0.1488\:moles\:of\:CuCl_2}{0.350L\:of\:solution}=0.425M\:CuCl_2$

3. 3.35g CaCO₃ in 500ml

Element number of atoms atomic mass TOTAL

Ca 1 x 40.08g/mol = 40.08g/mol

C 1 x 12.01g/mol = 12.01g/mol

O 3 x 16.00g/mol = 48.00g/mol

100.09g/mol

$3.35g\:of\:CaCO_3\times\dfrac{1\:mole\:of\:CaCO_3}{100.09\:g\:of\:CaCO_3}=0.0335\:moles\:of\:CaCO_3$

500ml = 0.5L

$\dfrac{0.0335\:moles\:of\:CaCO_3}{0.5L}=0.067M\:CaCo_3$

4 0

3 years ago

To help you just Incase you need it

Lerok [7]

Thanks thank u so much

5 0

3 years ago

Describe the effect an increase in i, the interest rate applied to the present value, has on the monthly payment P in the formul

VARVARA [1.3K]

Answer:

B. An increase in i, the interest rate, will create an increase in P, the monthly payment.

Step-by-step explanation:

We have the formula for the monthly payment as,

$P=\frac{i \times PV}{1-(1+i)^{-n} }$ ,

where P = monthly payment, i = rate of interest, PV = present value and n = time period.

Now, as i increase we get that (1+i) increases and so $(1+i)^{n}$ increases.

This gives us that, $\frac{1}{(1+i)^{n} }$ decreases and so $1-\frac{1}{(1+i)^{n}}$ decreases

Therefore, $\frac{1}{1-(1+i)^{-n} }$ increases.

So, we get that as i increases , the value of P will increase.

Hence, option B is correct.

4 0

3 years ago

Read 2 more answers

What is the median of the lower half of the data 5,8,10,1,7,6,15,8,6

Degger [83]

I’m think it is the median

5 0

3 years ago

Read 2 more answers

I will give you BRAINLEST if you answer

mafiozo [28]

Answer:

A)

Step-by-step explanation:

the top one

8 0

3 years ago