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3 years ago

PLEASE HELP ASAP WITH THESE THREE!

Chemistry

1 answer:

Furkat [3]3 years ago

5 0

Answer:

001: 71.2 g per ML

002: Consider a block of marble that occupies 287 cm3 and has a mass of 869 g.

a. What is its density?

d=m/v=869g/287 cm3=3.03 g/cm3

b. Will it float in:

H2O(l)? Why or why not?

Nope! Density > 1.00 g/mL

Hg(l) (density=13.55 g/mL)? Why or why not?

Yep! Density < 13.55 g/mL

003:

Use the given functions to set up and simplify

0.309 cm3

g/cm3=g/cm3

0.309cm3=g/cm3

Explanation:

http://childschemistry.weebly.com/uploads/7/8/6/3/78638416/density_key.pdf

Hope this helps I tried my best

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n the laboratory, two forms of sodium phosphate will be available (the monobasic monohydrate NaH2PO4·H2O, F.W. = 137.99 g/mol, a

const2013 [10]

Answer:

The compound you will use is the Dibasic phosphate

Explanation:

Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,

In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.

To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.

6 0

3 years ago

How is the preodictable arranged by atomic mass, coloms , and rows

bearhunter [10]

answer:an arrangement of elements in columns, based on a set of properties that repeat from row to row

Explanation:

hope this helps love <3

5 0

3 years ago

Read 2 more answers

Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

Goryan [66]

Answer: 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$

mass of $Fe_2(SO_4)_3$ = $moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g$

Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

3 0

3 years ago

For the following reaction, KcKc = 255 at 1000 KK.

bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.1550 0.173 0

At eqm. (0.1550-x) (0.173-x) x

As we are given:

$K_c=255$

The expression for equilibrium constant is:

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get:

$255=\frac{(x)}{(0.1550-x)\times (0.173-x)}$

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

5 0

3 years ago

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$