First answer will be brainliest!!

Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your

Minchanka [31]

When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

= 85 J

5 0

2 years ago

Five gases combined in a gas cylinder have the following partial pressures: 3.00 atm (N2), 1.80 atm (O2), 0.29 atm (Ar), 0.18 at

levacccp [35]

In order to solve the total pressure that is exerted by the gases, we need to use the Dalton's Law of Partial pressures. These are the calculations that you need to find out the total amount of pressure exerted to the gases:

3.00atm (N2) + 1.80atm (O2) + 0.29atm (Ar) + 0.18atm (He) + 0.10atm (H),
add up all of that, and the answer would turn out to be: 5.37atm.

8 0

2 years ago

Read 2 more answers

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

7 0

3 years ago

Solve question 4a least reactive to most reactive

denis23 [38]

Nickel (II) oxide, iron (III) oxide, chromium (III) oxide, magnesium oxide

7 0

2 years ago

Do equal volumes of the same gas at the same temperature and pressure contain equal numbers of molecules (number of moles)?

tresset_1 [31]

Answer:Therefore, there must be a direct relationship between these volumes of gases and the number of molecules they contain. Avogadro's law says that: Equal volumes of different gaseous substances, measured under the same pressure and temperature conditions, contain the same number of molecules.

Explanation:I hope it works for you.

4 0

3 years ago