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pishuonlain [190]
3 years ago
6

All organisms have the same number of cellsA. trueB. false​

Chemistry
2 answers:
denis-greek [22]3 years ago
8 0
The answer is B because not all organisms have the same number of cells
MakcuM [25]3 years ago
4 0

B. false is the answer of this question

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You have a 15.0 gram sample of gold at 20.0°C. How much heat does it take to raise the temperature to 100.0°C?
Nadusha1986 [10]

Answer:

=154.8 J

Explanation:

The rise in temperature is contributed by the change in temperature.

Change in enthalpy = MC∅,  where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.

Change in temperature = 100.0°C-20.0°C=80°C

ΔH=MC∅

The specific heat capacity of gold= 0.129 J/g°C

ΔH= 15.0g×0.129J/g°C×80°C

=154.8 J

7 0
3 years ago
Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr.
hram777 [196]

Answer:

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL

5 0
2 years ago
Three quantum numbers for an electron in a hydrogen atom in a certain state are
OLga [1]
It is option A 4s cause n=4, l=1 this is 4s orbital
6 0
2 years ago
At the equilibrium point in the decomposition of phosphorus pentachloride to chlorine and phosphorus trichloride, the following
Hitman42 [59]

Answer: K_{eq} for the reaction is 5.55

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

                        PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)          

At eqm. conc.   (0.010) M     (0.15)  M   (0.37) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}

Now put all the given values in this expression, we get :

K_c=\frac{(0.37)\times (0.15)}{(0.010)}

K_c=5.55

Thus the K_{eq} for the reaction is 5.55

5 0
3 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
2 years ago
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