the answer choices for all are shifts to the right, shifts to the left, and no change I need help plsssss

Calculate the mass of oxygen in 30 g of CH NH COOH?

Naya [18.7K]

Molar mass of CH2NH2COOH - 75

Given mass of CH2NH2COOH - 30

Moles of CH2NH2COOH = Given mass/ Molar mass

moles of CH2NH2COOH = 30/75 = 0.4 mol

One mole of CH2NH2COOH contains 32 gram of oxygen

0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen

Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!

7 0

3 years ago

A sample of neon gas is at a temperature of 60°C and has a volume of 300 cm. What is the volume of the gas if it is heated to 10

elena55 [62]

Answer:

336.04 cm

Explanation:

T1 = 60+273 = 333K

T2 = 100+273 = 373K

V2 = (V1×T2)/T1

V2 = 300×373/333

V2 = 336.04 cm

6 0

3 years ago

Chemistry is the definition

KIM [24]

Answer:

the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.

Explanation:

Hope this helps. :)

8 0

4 years ago

Calculate each of the following quantities. (a) Mass (g) of solute in 175.4 mL of 0.267 M calcium acetate WebAssign will check y

ss7ja [257]

Answer:

a) 7.40 grams

b) 0.213 M

c) 102 moles

Explanation:

(a) Mass (g) of solute in 175.4 mL of 0.267 M calcium acetate

Step 1: Data given

Volume = 175.4 mL = 0.1754 L

Molarity = 0.267 M

Molar mass = 158.17 g/mol

Step 2: Calculate moles

Moles = molarity * volume

Moles = 0.267 M * 0.1754 L

Moles = 0.0468 moles

Step 3: Calculate mass

Mass = 0.0468 moles * 158.17 g/mol

<u>Mass = 7.40 grams</u>

b) Molarity of 597 mL solution containing 21.1 g of potassium iodide

Step 1: Data given

Volume = 597 mL = 0.597 L

Mass = 21.1 grams

Molar mass KI = 166.0 g/mol

Step 2: Calculate moles KI

Moles KI = 21.1 grams / 166.0 g/mol

Moles KI = 0.127 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.127 moles / 0.597 L

Molarity = 0.213 M

(c) Amount (mol) of solute in 145.6 L of 0.703 M sodium cyanide

Step 1: Data given

Volume = 145.6 L

Molarity = 0.703 M

Step 2: Calculate moles

moles = molarity * volume

Moles = 0.703 M * 145.6 L

Moles = 102 moles

8 0

3 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

6 0

3 years ago