Need help asap will give brainly

Vesna [10]

From Q = mcΔΤ, the specific heat capacity, c, of the metal that was cooled is c = Q/mΔT = (-769 J)/(46.4 g)(30.0 °C - 101.0 °C) = 0.233 J/g °C. From the table, it appears that this is the specific heat capacity of silver. So, the metal is most like silver.

Note: The value for Q was written as a negative value in the equation as heat energy was given off by the metal when the metal was cooled (from the metal’s point of view, it’s losing heat energy).

4 0

3 years ago

What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N

Citrus2011 [14]

<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

Step 1: Define

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

Step 2: Identify Conversions

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

Step 3: Stoich

[DA] Set up: $\displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.68873 \ g \ Mg_3N_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0

2 years ago

What is the main difference between a generator and an electric motor? _______

garri49 [273]

Answer:

A generator produces electricity and an electric motor consumes electricity

4 0

2 years ago

Read 2 more answers

Convert 512 kilograms to milligrams. 0.000512 0.512 512,000 512,000,000

olga55 [171]

The answer is D! 512000000

5 0

3 years ago

Read 2 more answers

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

This last equation is the Clapeyron equation.

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago