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Free_Kalibri [48]

3 years ago

14

How did rutherford know that the nucleus was positively charged?

1 answer:

AleksAgata [21]3 years ago

3 0

'cause alphe-particle which was +ve charge, get repulsion from the atom, so he deducted that.......

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Calculate the energy in joules and calories required to heat 25.0 g of water from 12.5C to 25.7C

Alex Ar [27]

The heat (Q) required to raise the temp of a substance is:<span>Q=m∗Cp∗ΔT</span><span> where m is the mass of the object (25.0g in this case), Cp is the specific heat capacity of the substance (for water Cp = 1.00cal/gC, or 4.18J/gC,
and Dt is the change in temp.
You'll have to solve this twice, once with the Cp in calories, and once with the Cp in joules.
</span><span>1380.72 Joules</span>

8 0

3 years ago

The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af

Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K$

Putting values in above equation, we get:

$\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa$

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

4 0

3 years ago

3 kelebihan watak utama dalam novel meniti impian

In-s [12.5K]

I honestly don’t know sorry

7 0

3 years ago

Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn

MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0

3 years ago

Read 2 more answers

What is the pH of pure water at 40.0Â°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233

Sergeeva-Olga [200]

Answer:

PH= 6.767 (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

$[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\$

at 40ºC

$[H_{3}O^{+} ] = [OH^{-} ]$

$[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}$

$[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M$

$PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767$

7 0

3 years ago