fire transfers heat by ___, meaning some heat rises, and ___, meaning some heat goes out in every direction.

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

3 0

3 years ago

Which of the following are properties of metals? Select all that apply.

earnstyle [38]

Answer:

All of them

Explanation:

All of those are properties of metals

3 0

3 years ago

Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

Solubility in 0.0120 M CoBr₂ (S')

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0

3 years ago

Write a hypothesis statement about the relationship between the number of mushrooms and the concentration of heavy metals in the

Anna007 [38]

Answer:

Write a hypothesis statement about the relationship between the number of mushrooms and the concentration of heavy metals in the soil ...

8 0

2 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

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

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago