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Archy [21]
4 years ago
10

The conjugate base of H2CO3 is ___________

Chemistry
2 answers:
olasank [31]4 years ago
7 0

Answer:

HCO₃

Explanation:

SOVA2 [1]4 years ago
4 0

Base loses H⁺ ion

HCO₃⁻

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Satellite photos of Neptune and Uranus show that both planets have a bluish color. Which gas in their atmospheres gives them thi
Shtirlitz [24]

Answer:

Methane

Explanation:

5 0
3 years ago
Match the time frame listed on the left with the characteristic listed on the
SVEN [57.7K]

mesozoic era - age of the dinosaurs

end of the paleozoic era - most of earth a desert

end of the mesozoic era - a layer of iridium deposited all over earth

5 0
3 years ago
15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °C. Calculate the equilibrium tempe
elena-14-01-66 [18.8K]

Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

62.76T2 - 627.6 = -627.6T2 + 49329.36

690.36T2 = 49956.96

T2 = 72.4 °C

The equilibrium temperature of the coffee is 72.4 °C

7 0
3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
Tris base has a molecular weight of 121 g/mol. How many grams of tris base would you need to make 250ml of a 200mM SOLUTION
STatiana [176]

Answer:

6.05 g

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

From the question ,

M = 200mM

Since,

1 mM = 10⁻³ M

M = 200 * 10⁻³ M

V = 250 mL

Since,

1 mL = 10⁻³ L

V = 250 * 10⁻³ L

The moles can be calculated , by using the above relation,

M = n / V  

Putting the respective values ,

200 * 10⁻³ M = n / 250 * 10⁻³ L

n = 0.05 mol

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

m = 121 g/mol

n = 0.05 mol ( calculated above )

The mass of tri base can be calculated by using the above equation ,

n = w / m  

Putting the respective values ,

0.05 mol = w / 121 g/mol

w = 0.05 mol * 121 g/mol

w = 6.05 g

3 0
4 years ago
Read 2 more answers
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