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2 years ago

If a gas is measured to have a pressure of 1.2 atm at 227 K, and its origina

Chemistry

1 answer:

ICE Princess25 [194]2 years ago

7 0

Answer:

1.05

Explanation:

P1 over t1 is equal to P2 over T2, then solve for X

X/199= 1.2/227

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Please answer all of the questions

anzhelika [568]

Here's your ans

Explanation:

1-1.50

2-d

3-c

4-b

5-d

3 0

2 years ago

Draw Conclusions does the F2 generation support that lack of wings is a genetic trait? Explain why or why not

Dmitrij [34]

Answer:

An F1 offspring could produce four types of gametes, RY, Ry, rY, and ry. The F2 generation supports the independent-assortment model and refutes the linkage model.

Explanation:

3 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

2 years ago

A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure

GenaCL600 [577]

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

Best regards!

8 0

2 years ago

24 grams of iron is added to a graduated

Vanyuwa [196]

Answer:

d = 8 g/mL

Explanation:

Given data:

Mass of metal = 24 g

Volume of eater = 45 mL

Volume of water + metal = 48 mL

Density of iron metal = ?

Solution:

Volume of metal:

Volume of metal = volume of water+ metal - volume of water

Volume of metal = 48 mL - 45 mL

Volume of metal = 3 mL

Density of metal:

d = m/v

d = 24 g/ 3 mL

d = 8 g/mL

4 0

2 years ago