Help please I’m being timed!

A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I

murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0

3 years ago

Pls help me for this one :) is hard for me thx

zvonat [6]

Answer:

protein

Explanation:

that the answer

8 0

3 years ago

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A 100.0 mL sample of a 0.200 M aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 M aqueous solution of BaCl2. The mixt

stepan [7]

The correct answer is 2.53 g of precipitate, BaCrO4.

5 0

3 years ago

Titanium and chlorine react to form titanium(IV) chloride, like this:

leva [86]

Answer:

the value of equilibrium constant for the reaction is 8.5 * 10⁷

Explanation:

Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)

equilibrium constant Kc = $\frac{1}{[Cl_2]^2}$

Given that,

We are given:

Equilibrium amount of titanium = 2.93 g

Equilibrium amount of titanium tetrachloride = 2.02 g

Equilibrium amount of chlorine gas = 1.67 g

We calculate the No of mole = mass / molar mass

mass of chlorine gas = 1.67 g

Molar mass of chlorine gas = 71 g/mol

mole of chlorine = 1.67 / 71

= 7.0L

Concentration of chlorine is = no of mole / volume

= 0.024 / 7

= 3.43 * 10⁻³M

equilibrium constant Kc = $\frac{1}{[Cl_2]^2}$

= $\frac{1}{[3.43 * 10^-^3]^2}$

= 8.5 * 10⁷

4 0

4 years ago

What is the mass, in grams, 6.5 mol of sodium nitrate (nano3)? please show all work?

ladessa [460]

Mass is the property of a physical body and the resistance to acceleration when a net force is applied on the body.
The atomic mass of sodium (Na) is = 22.98
The atomic mass of nitrate (N) is = 14.00
The atomic mass of oxygen (O) is = 15.99
The sodium nitrate (NaNO3) consists of the atomic masses of Na+N+(O)3 = 85 grams
Therefore, the mass of 6.5 mol of sodium nitrate is = 6.5 * 1 mol of NaNO3
= 6.5 * (85)
= 552.50 grams

6 0

3 years ago

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