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klemol [59]
2 years ago
5

Someone help pls :,)

Chemistry
2 answers:
Snowcat [4.5K]2 years ago
5 0

Answer:D

Explanation:

Cause I literally just did this

Ulleksa [173]2 years ago
4 0
Ionic compound is NaOH do letter D should be correct.
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A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming
ahrayia [7]

Answer:

sorry but i cant

Explanation:

8 0
3 years ago
A chemist has some 40% acid solution, some 60% acid solution, and a wholebunch of free time. How many liters of each should be u
TiliK225 [7]

Answer:

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

Explanation:

Let volume of the 40% acid solution be x.

Let volume of the 60% acid solution be y.

Volume of solution formed after mixing both solution = 40 L

x + y = 40 L..[1]

Volume of acid 40% solution = 40% of x= 0.4x

Volume of acid 60% solution = 60% of y= 0.6y

Volume of acid formed = 45% of 40 L = \frac{45}{100}\times 40=18L

0.4x+0.6y=18 L..[2]

Solving [1] and [2]

x = 30 L  ,   y = 10 L

30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.

8 0
3 years ago
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
3 years ago
How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?
saw5 [17]

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

5 0
2 years ago
zn (s) 2 hcl (aq) -----&gt; zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0
nadya68 [22]

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn

Read more about zinc;

brainly.com/question/28880469

#SPJ4

3 0
1 year ago
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