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3 years ago

HELP! WILL GIVE BRAINLIEST DUE SOON

2 answers:

5 0

Dome Mountains are formed from hot molten material (magma) rising from the Earth's mantle into the crust that pushes overlying sedimentary rock layers upward to form a “dome” shape.
So it’s D.

Flauer [41]3 years ago

4 0

Answer:

I am not the best to answer this but I think it would be B

Explanation:

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Which mineral might scratch the mineral fluorite but would not scratch the mineral amphibole

morpeh [17]

Gypsum is the mineral

8 0

4 years ago

Read 2 more answers

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

How many grams of Al are needed to react with 63.0 g of FeO

Tema [17]

The amount of Al required will be 15.77 grams

<h3>Stoichiometric problem</h3>

First, the equation of the reaction:

$2Al +3 FeO -- > Al_2O_3 + 3Fe$

The mole ratio is 2:3.

Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles

Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles

Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

4 0

2 years ago

Read 2 more answers

Calculate the mass (in grams) of methylene bluecrystals that you must weigh in order to make 100.0mL of 1.25 × 10-5mol/L methyle

mario62 [17]

<u>Answer:</u> The mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $1.25\times 10^{-5}M$

Molar mass of methylene blue = 319.85 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$1.25\times 10^{-5}M=\frac{\text{Mass of methylene blue}\times 1000}{319.85\times 100.0}\\\\\text{Mass of methylene blue}=\frac{1.25\times 10^{-5}\times 319.85\times 100.0}{1000}=3.99\times 10^{-4}g$

Hence, the mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

4 0

3 years ago

Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skelet

saw5 [17]

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

5 0

4 years ago