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2 years ago

What music was made in 1920

Chemistry

2 answers:

Setler79 [48]2 years ago

7 0

Jazz, blues, ragtime :)
Can I have brainliest?

Grace [21]2 years ago

4 0

Answer:

Jazz, Blues, Swing, Dance Band, And Ragtime

Explanation:

I hope this helps

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PLEEASE HELP TIMED WILL GIVE BRAINLIEST TO CORRECT ANSWER NLY PLEASEEEE HELPPPP

Arturiano [62]

Answer:

1) 1.15 mol

2) M=0.45

3) 22.5 mL

4) 6.25 mL

Explanation:

550 mL= 0.55 L

M= mol solute/ L solution

mol solute= M * L solution

mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute

155 mL = 0.155 L

80 g -> 1 mol NH4NO3

5.61 g -> x

x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3

M=(0.07 mol NH4NO3)/0.155 L M=0.45

3) M1V1=M2V2

V1= M2V2/M1

V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL

4) M1V1=M2V2

V1= M2V2/M1

V1= (0.25 M * 0.45 L)/ 18.0 M

V1=6.25 x 10^-3 L = 6.25 mL

8 0

2 years ago

Read 2 more answers

What is true about an element

stich3 [128]

Answer:

made from ine atom

Explanation:

one atom

5 0

2 years ago

Li + HNO3 > LiNO3 + H2 how do I balance it? Show work pls

azamat

Answer: The balanced equation is $2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$ .

Explanation:

The given reaction equation is as follows.

$Li + HNO_{3} \rightarrow LiNO_{3} + H_{2}$

Number of atoms present on reactant side are as follows.

Li = 1
H = 1
$NO_{3}$ = 1

Number of atoms present on product side are as follows.

Li = 1
H = 2
$NO_{3}$ = 1

To balance this equation, multiply Li by 2 and $HNO_{3}$ by 2 on reactant side. Also, multiply $LiNO_{3}$ by 2 on product side.

Hence, the equation can be rewritten as follows.

$2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$

Now, number of atoms present on reactant side are as follows.

Li = 2
H = 2
$NO_{3}$ = 2

Number of atoms present on product side are as follows.

Li = 2
H = 2
$NO_{3}$ = 2

As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.

Thus, we can conclude that the balanced equation is $2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$ .

3 0

2 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

4 0

2 years ago

To make sure that the white powder was all sodium chloride and not mixed with sodium hydrogen carbonate, would you need to add a

TiliK225 [7]

Answer: i hope you thought this was an actual answer

Explanation:

6 0

2 years ago