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sergey [27]
3 years ago
10

Help pls very simple

Chemistry
1 answer:
nalin [4]3 years ago
7 0
What equipment is generally used to make lyophilized medications suitable for administering to the patient
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What are the types of precipitation?
s344n2d4d5 [400]

Answer:

Rain,Hail,Snow,sleet. Hope this helped you out!

8 0
2 years ago
Copper (II) sulfate pentahydrate may be heated to drive off the water of hydration. If 5 g of water are produced, what was the o
____ [38]

Answer:

13.9g of CuSO4•5H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CuSO4•5H2O → CuSO4 + 5H2O

Next, we shall determine the mass of CuSO4•5H2O heated and the mass of H2O produced from the balanced equation.

This is illustrated below

Molar mass of CuSO4• 5H2O = 63.5 + 32 + (16x4) + 5(2x1 + 16)

= 63.5 + 32 + 64 + 5(18) = 249.5g/mol

Mass of CuSO4•5H2O from the balanced equation = 1 x 249.5 = 249.5g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 5 x 18 = 90g

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Now, we can determine the mass of CuSO4•5H2O needed to produce 5g of H2O. This can be achieved as shown below:

From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

Therefore, Xg of CuSO4•5H2O will produce 5g of H2O i.e

Xg of CuSO4•5H2O = (249.5 x 5)/90

Xg of CuSO4•5H2O = 13.9g

Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.

6 0
3 years ago
Which of the following are examples of matter?a. heat b. sunlight c. water d. grass e. air
OverLord2011 [107]
Sunlight? air? i’m not sure they could all be matter..
7 0
4 years ago
A man drove his car through the Nevada desert one day when the temperature was over 100°F. When he started the trip, the gas sta
Slav-nsk [51]
<span>The temperature in the tire increased, causing an increased tire pressure. :D</span>
7 0
3 years ago
1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
Ivanshal [37]

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ

Regards!

7 0
3 years ago
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