All categories

3 years ago

What volume of 18.5 M nitric acid would be needed to make 25.0 mL of 3.0M?

1 answer:

5 0

Answer:What volume of 9.00 M nitric acid is needed to make 6.50 L of 1.25 M solution? 903 mL What molarity should the stock solution be if you want to dilute 25.0 mL to 2.00 L and have the final concentration be 0.103 M?

Explanation:

You might be interested in

Heating a substance____the speed at which molecules move

guapka [62]

Answer:

Heating a substance increases the speed at which molecules move.

Explanation:

4 0

3 years ago

Read 2 more answers

Which of the following are parts of the excretory system? *

Kay [80]

Answer:

kidney, bladder, rectum, urethera

Explanation:

5 0

3 years ago

Helllpppp!!!!!! Please

pav-90 [236]

Answer: (C) Statements (i) and (iii)

Explanation: According to byjus.com, group VII elements are known as Halogens.

Not only that, but bbc.co.uk says " Atoms of group 7 elements all have seven electrons in their outer shell. This means that the halogens all have similar chemical reactions ."

It may just be (b) though as these are chemical reactions.

3 0

2 years ago

Explain how you would decide which represents a greater mass: 3.25 x 1022 atoms of Ca or 2.45 grams of Mg?

elena55 [62]

Answer:

You need to convert the number of atoms of Ca into mass in grams, using Avogadro's number and the atomic mass of Ca.

Explanation:

The amount of matter is measured in grams. Thus, you need to convert the number of atoms of Ca (calcium) into mass to compare with 2.45 grams of Mg.

To convert the atoms of calcium into mass, you divide by Avogadro's number, to obtain the number of moles of atoms, and then divide by the atomic mass of calcium.

1. Number of moles, n

$n=\dfrac{\text{number of atoms}}{\text{Avogadro's number}}\\ \\ \\ n=\dfrac{3.25\times 10^{22}}{6.022\times 10^{23}}=0.053969mol$

2. Mass

mass = number of moles × atomic mass

mass = 0.053969mol × 40.078g/mol = 2.16g

Then, 2.45 g of Mg represent a greaer mass than the 3.25 × 10²² atoms of Ca.

4 0

3 years ago

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

7 0

3 years ago