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lbvjy [14]
3 years ago
15

Rank these gas samples according to pressure

Chemistry
1 answer:
stiv31 [10]3 years ago
4 0
I honestly do not know the answer. sorry
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What is the percent cr, by mass, in the steel sample? express your answer numerically as a percentage?
djverab [1.8K]
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−

Assuming all the Cr is contained in the BaCrO4 at the end. 
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
8 0
3 years ago
Read 2 more answers
Need help on question 3 plz<br> URGENT HELP PLZ
Law Incorporation [45]
Alloys are the homogeneous mixture of metals and non metals.... they are used in making cars and jewellery to make them more featurable,,, means that they would possess high regidity to the destruction .... both features of metal and nonmetal
5 0
3 years ago
How do you find the amount of moles is .032 grams of water and whats the answer
masha68 [24]

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

  • 2 Hydrogen atoms (subscript of 2)
  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
  • Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

6 0
3 years ago
Convert 10kg⋅cm/s^2 to newtons
Solnce55 [7]

Answer:

one newton

Explanation:

tell me if it wrong

3 0
3 years ago
Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is described by the following thermoche
vredina [299]

Answer:

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.

Explanation:

4C_2H_5O_2N(s) + 9O_2(g)\rightarrow 8CO_2(g) + 10H_2O(l) + 2N_2(g) ,\Delta H^{rxn} =-3896 kJ/mol

Standard enthalpy of formation of oxygen gas= \Delta H_{f,O_2}=0

Standard enthalpy of formation of carbon dioxide= \Delta H_{f,CO_2}=-393.5 kJ/mol

Standard enthalpy of formation of water = \Delta H_{f,H_2O}=-285.8 kJ/mol

Standard enthalpy of formation of nitrogen gas= \Delta H_{f,N_2}=0

Standard enthalpy of formation of glycine = \Delta H_{f,gly}=?

Enthalpy of the reaction = H_{rxn}=-3896 kJ/mol

H_{rxn} =

=8\times \Delta H_{f,CO_2} +10\times \Delta H_{f,H_2O}+2\times \Delta H_{f,N_2} - (4\times \Delta H_{f,gly}+9\times \Delta H_{f,O_2})

-3896 kJ/mol=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+0 - (4\times \Delta H_{f,gly} +0)

On rearranging :

4\times \Delta H_{f,gly}=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+ 3896 kJ/mol

\Delta H_{f,gly}=\frac{-2149 kJ/mol}{4}=-537.25 kJ/mol

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.

6 0
3 years ago
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