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Goshia [24]
3 years ago
8

How much heat is required to raise the temperature of 835g of water from 35.0°C to 65.0°C?

Chemistry
1 answer:
Anna007 [38]3 years ago
3 0

Answer:

25,050 calories.

Explanation:

A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.

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Can light bend around corners to reach an object
Sholpan [36]

Answer: Yes, light can bend around corners. In fact, light always bends around corners to some extent.

Explanation:This is a basic property of light and all other waves. ... The ability of light to bend around corners is also known as "diffraction".

7 0
3 years ago
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Which of the following describes the arrangement of valence electrons in a bond between H and F?
Sloan [31]
C; The Valence electrons spend more time around the atom of F
6 0
3 years ago
Can someone plz help me with this plz?????
Otrada [13]

Answer:

We can't see the options so we don't know what we can put

Explanation:

4 0
3 years ago
The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, an
Andre45 [30]

The height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

<h3>What is Pascal? </h3>

Pascal is defined as the force per unit area. It expressed in Newton pr square meter of area.

1 Pa = 1 N / m²

Pressure = force / Area

According to the question, the expression of force is as given below

F = mg

where,

F is the force,

m is the mass,

g is equal to the acceleration due to the gravity

Now, the area of the atmosphere is 1 m².The pressure is 1 atm. Pressure in Pascal.

1 atm = 1.01325 × 10 5pa

Therefore, the expression for pascal become as follows.

1.01325 × 10 5 pa = mg /area

1.01325 × 10 5 pa = m × 9.81 m/s² / 1 m²

M = 1.01325 × 10 5 pa × 1 m² / 9.81 m² × 1 Nm -² /1 pa × 1 kg m-² / 1 N

1.03 × 10 ⁴ kg

Given,

The density is 22.6 g /mL , pressure is 1 atm, and area is 1 m²

The relation between density and pressure can be given as follows.

P = hpg… … …(1 )

were , h is the height of the column

p is the density.

Hpg = 1.01325 × 10 5 pa × 1 N/m² /1 pa

H = 1.01325 × 10 5 N/m² / pg × 1 kg ms-² / 1 N

= 1.01325 × 10 5 kg m-¹ s -² / 22.6 g mL -1 × 1 kg/ 10 ³ g × 1 mL / 10 -6 m ³ × 9.81 m s- ²

= 0.457 m

Therefore, the height of the column is 0.457 m.

Thus, we concluded that the height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

learn more about density:

brainly.com/question/952755

#SPJ4

7 0
2 years ago
s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g
Elenna [48]

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

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express  your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Now put all the given values in above equation, we get:

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

Hence, the volume of hydrogen gas that will be collected is 1.85 L

8 0
3 years ago
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