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3 years ago

Help needed ASAP plz

Chemistry

1 answer:

Sati [7]3 years ago

7 0

Answer:

6 cucumberslices.

Explanation:

You can make a maximum of 4 sandwiches. 8 slices of tomato and 2 needed for each. 8/4=2. You need 4 slices of cucumber per sandwich and have 22. the maximum of sandwiches we can make is 4 so 4×4=16. 22-16=6 leftover slices of cucumber.

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A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0

3 years ago

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

3 0

3 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

3 years ago

An atom with a high metallic character usually has a ____ electron affinity and a ____ atomic radius.

lianna [129]

Low electron affinity and large atomic radius

Metallic character decreases across a period (from left to right) and increases down a group.

Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius. Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius.

8 0

3 years ago

SCIENCE:how can i model a ice cream maker melting!?

lord [1]

Answer:

BURN IT ALIVE MUHAHAHAHAHA

Explanation:

6 0

3 years ago