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3 years ago

SORRY BUGGING ALL YALL AGAIN NEED HELP DUMB SORRY IF YOUR READING THIS U KNOW WHO U R D

Chemistry

2 answers:

natita [175]3 years ago

8 0

Answer:

hope this helps

please like and Mark as brainliest

Explanation:

reduce eliminate
interdependent
mutually beneficial, competitive and predatory

garik1379 [7]3 years ago

7 0

Answer:

Dont think your dumb because your not you simply just dont understand the question

Explanation:

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30,000 J of heat are added to 23.0 kg of steel to reach a final temperature of 140

kolbaska11 [484]

The initial temperature is 137.34 °C.

Explanation:

As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.

Q = mcΔT

So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.

ΔT = $\frac{Q}{mc}$

ΔT = $\frac{30000}{23 \times 490} = \frac{30000}{11270} =2.66$

Since the difference in temperature is 2.66, then the initial temperature will be

Final temperature - Initial temperature = Change in temperature

140-Initial temperature = 2.66

Initial temperature = 140-2.66 = 137.34 °C

Thus, the initial temperature is 137.34 °C.

7 0

3 years ago

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Nataly_w [17]

Answer: 1.61 x 10⁴ kPa

Dalton's law states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture. According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt), which is calculated by adding the moles of all its components:

xi = $\frac{n_{i} }{n_{t} }$

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ = 10 g x $\frac{1 mol}{31.998 g}$ = 0.313 mol

n₂ = 50 g x $\frac{1 mol}{8.005 g}$ = 6.246 mol

n₃ = 40 g x $\frac{1 mol}{28.013 g}$ = 1.428 mol

Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ = $\frac{6.246 mol}{7.987 mol}$ = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

So, the partial pressure of helium if a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa is 1.61 x 10⁴ kPa

5 0

3 years ago

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;

$Normality=molarity \times n-factor$

but in case of NaOH and HCl n-factor is 1 for each.

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

What kind of chemical reaction is Al2+S3

Vikki [24]

Answer:

Aluminum sulfide or aluminium sulphide is a chemical compound with the formula Al2S3.

Explanation:

4 0

3 years ago

Which of the following has kinetic energy?

shepuryov [24]

Answer:

moving arrow

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4 0

2 years ago

Read 2 more answers