(2 points) How many grams of K2SO4 are present in 25.0 mL of 7.00 % (m/v) solution?
Show your work. No work = no credit.

Answers

Answer 1

Mass of K₂SO₄ : 1.75 g

Further explanation

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

volume of solution = 25 ml

% (m/v)=7%

mass of K₂SO₄ :

[tex]\tt \%m/v=\dfrac{mass~of~solute}{volume~of~solution}\times 100\%\\\\7\%=\dfrac{mass~K_SO_4}{25}\times 100\%\\\\mass~K_2SO_4=7\times 25\div 100=1.75~g[/tex]


Related Questions

A nurse practitioner orders Medrol to be given 1.4 mg/kg of body weight. Medrol is an anti-inflammatory administered as an intramuscular injection. If a child weighs 71.6 lb and the available stock of Medrol is 20. mg/mL, how many milliliters does the nurse administer to the child? Express your answer to two significant figures and include the appropriate units.​

Answers

The nurse give 2.3 ml Medrol to the child

Further explanation

Mass is one of the principal quantities, which is related to the matter in the object  

The main mass unit consists of 7 units

kg, hg, dag, g, dg, cg, mg

Conversion :

1 Lb=0,453592 kg

For 71.6 lb :

[tex]\tt 71.6\times 0.453592=32.48~kg[/tex]

Medrol to be given 1.4 mg/kg of body weight, so amount for the child :

[tex]\tt 32.48~kg\times 1.4~mg/kg=45.472~mg[/tex]

The available stock of Medrol is 20. mg/mL, so the volume to the child :

[tex]\tt \dfrac{45.472}{20}=2.2736\approx2.3~ml[/tex]

Elements of the same subgroup have the same number of electrons on the outer layer. Am i right?​

Answers

Answer:

yes

Explanation:

because they have similar chemical properties

Plzz helppppppppppppp

Answers

Answer:

i got you stop doing work  and forget about this one class you will still pass trust me i did it it 8th grade and still made it to 9th but today im not even in school so just do better in your other classes

Explanation:

With ionic compounds, the ion charge is the same as the ___ number.
A. methylation
B. anion
C. cation
D. oxidation

Answers

Answer:

D. Oxidation

Explanation:

The oxidation state of a pure element is always equal to zero and the oxidation state for a pure ion is equivalent to its ionic charge.

How much ice could be melted at 0​°​C if 5200 joules of heat were added?

Answers

Answer:

0.02kg

Explanation:

Given parameters:

Amount of heat  = 5200J

Unknown:

Mass of ice that would be melted at 0°C = ?

Solution:

To solve this problem, use the expression below;

        H  = mL

H is the heat

m is the mass

L is the latent heat of fusion of ice  =  3.33 x 10⁵ J/kg.

 Insert the parameters and solve for m;

           5200  = m x 3.3 x 10⁵

             m  = [tex]\frac{5200}{3.33 x 10^{5} }[/tex]   = 0.02kg

     

What term describes the state of matter in which the attractive forces between particles are the greatest?


liquid

solid

gas

metal

Answers

It would probably be C

Answer:

I think it would be b

Explanation:

Does it matter which of the two sp3sp3 hybrid orbitals are used to hold the two nonbonding electron pairs

Answers

Answer:

No. All four hybrids are equivalent and the angles between them are all the same, so we can use any of the two to hold the nonbonding pairs.

Explanation:

Orbital hybridization has been the combination of the atomic orbitals for the formation of a new hybrid. No, it does not matter which orbitals are used to hold the nonbonding electron pairs.

What are hybrid orbitals?

Hybrid orbitals are said to be formed by the mixing of the atomic orbitals with different energy and geometrical shape that allows the understanding of the atomic bonding and molecular geometry of the compounds.

It includes sp, sp², sp³, sp³d, and sp³d² which have various arrangements, including linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. The hybrid orbitals are formed by the combination and overlapping of the s and p orbital.

It does not matter which sp³ orbital holds the two non-bonding electron pairs as all four hybrids have been known to have equivalent angles between them so it will not matter which hybrid has the non-bonded electrons in them.

Therefore, it will not matter which two sp³ hybrid holds the electron pairs that are not involved in bonding.

Learn more about orbital hybridization, here:

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Solve each of the following problems using ​dimensional analysis​.

Answers

Answer:

12ml should be the answer

Explanation:

This equation is balanced: N2+H2 -NH3

A. True

B. False​

Answers

The equation N2+H2= NH3 is balanced. This is because N2 + 3H^2 ➡️ 2NH^3

Which is an example of a beneficial mutation?

one that changes the color of a rabbit, allowing it to hide from predators.
one that results in lighter flower petal colors without changing the plant’s ability to reproduce
one that causes a person’s body to produce thick mucus that clogs the lungs
one that makes it easier for a corn plant to contract a disease

Answers

Answer:

A

Explanation:

Answer: a

Explanation: one that changes the color of a rabbit, allowing it to hide from predators.

1. How many moles of water molecules are there in 39 grams of Chlorine?

2. How many molecules are there in 39 grams of Gold?​

Answers

Answer:

1. There is no moles of water in 39g of chlorine.

2. 1.19x10²³ molecules of Au.

Explanation:

First, we need to remember that molar mass of a compound represents the mass of 1 mole (6.022x10²³ molecules) of molecules.

Molar mass of Chlorine, Cl₂ is 70.9g/mol

Molar mass of gold, Au, is 197g/mol

1. Moles of 39g of Cl₂ are:

39g Cl₂ * (1mol / 70.9g) = 0.55 moles moles of chlorine

But there is no moles of water in 39g of chlorine.

2. First, moles of Au are:

39g Au * (1mol / 197g) = 0.198 moles Au.

Molecules are:

0.198 moles Au * (6.022x10²³ molecules / 1 mol) =

1.19x10²³ molecules of Au

1. Which of the following statements describe cations? Select all that apply.
Positively charged ions.
Negatively charged ions.
They have lost electrons.
They have gained electrons.

Answers

Hey :)

Cations are positively charged ions and they have lost electrons

Hope this helps!

Ions can be made by single element or covalently bonded group of elements. The covalently bonded group of elements is called polyatomic ions or polyatomic atoms. The correct option are option A,C.

What is Ions?

Any species that contain charge whether it is positive charge or negative charge is called ions. The example of polyatomic ions are sulfate, phosphate, nitrate etc.

Cation is the species that loose electron and attain positive charge while anion is a species which gain electron and attains negative charge so when anion and cation combine in fixed ration the the overall charge of the molecule is zero that is molecule is neutral, the charge over cation and anion is also called oxidation state.

Therefore the correct option are option A,C.

To learn more about ions, here:

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To prepare 380 mL of 0.500 M NaCl (MW = 58.5 g/mol), how many grams of sodium chloride is required?

Answers

This answer did not come from me but here you go!
44 g NaCl
Explanation:
The problem provides you with the molarity and volume of the target solution, so your first step here will be to use this information to figure out how many moles of sodium chloride,
NaCl
, it must contain.
Once you know that, use the compound's molar mass to convert the number of moles to grams.
So, molarity is defined as the number of moles of solute per liter of solution. In your case, a
0.77 M
solution will contain
0.77
moles of sodium chloride, your solute, in
1.0 L
of solution.
Your target solution has a volume of
985
mL

1 L
10
3
mL
=
0.985 L
which means that it will contain
0.985
L solution

= 0.77 M

0.77 moles NaCl
1
L solution
=
0.75845 moles NaCl
Sodium chloride has a molar mass of
58.44 g mol

1
, which basically means that one mole of sodium chloride has a mass of
58.44 g
.
In your case,
0.75845
moles of sodium chloride will have a mass of
0.75845
moles NaCl

58.44 g
1
mole NaCl
=




¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
44 g
a
a

11.12 grams of sodium chloride is required to prepare 380 mL of 0.500 M NaCl (MW = 58.5 g/mol).

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of sodium chloride must first be calculated by multiplying the molarity by its volume.

Molarity = no. of moles ÷ volume

0.500M = n ÷ 0.380L

n = 0.5 × 0.380

n = 0.19mol

mass of NaCl = 58.5g/mol × 0.19mol

Mass of NaCl = 11.12g

Therefore, 11.12 grams of sodium chloride is required to prepare 380 mL of 0.500 M NaCl (MW = 58.5 g/mol).

Learn more about how to calculate mass at: https://brainly.com/question/14948089?referrer=searchResults

The isotope of carbon used in archaeological dating is 14^6C . How many protons, neutrons, and electrons does an atom of 14^6C have?

Answers

Answer:

6

8

6

Explanation:

 Isotope given:

                       ¹⁴₆C

In specie written as this;  

   Superscript  = Mass number

   Subscript  = Atomic number

To find the protons, it is the same as the atomic number;

                Protons  = Atomic number  = 6

Neutrons have no charges;

 Neutrons  = Mass number - Atomic number  =

  Neutrons  = 14 - 6  = 8

The number of electrons is the same as the atomic number = 6

. A 20.0 % by mass solution of phosphoric acid (H3PO4) in water has a density of 1.114 g/mL at 20°C. What is the molarity of this solution?
The molar mass of phosphoric acid (H3PO4) is 97.99 g.

Answers

Answer:

2.273M

Explanation:

What is the definition of molarity.

M = mols/L. So that's what we need to determine.

How much does a L weigh? That's

1.114 g/mL x 1000 mL = 1114 grams. Simple enough, eh?

How much of that 1114 g is H3PO4. It says it is 20% by mass, therefore, 1114 g x 0.20 = 222.8 g.

How many mols are there in 222.8 g H3PO4? That's mools = grams/molar mass = 222.8/98 = 2.273 mols.

The definition of M is what? M = mols/L. And you have 2.273 mol/L; that must be the molarity.

The formula is

density g/mL x 1000 mL x mass% x (1/molar mass) = M

1.114 x 1000 x 0.20 x (1/98) = 2.273 M.

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