The metal zirconium becomes superconducting at temperatures below 3.4000K.

Calculate the mass of water formed when 4g of methane is burnt in excess oxygen

wariber [46]

Answer:

Mass of water formed: 9g

Explanation:

Hope it helps you!

6 0

2 years ago

A gold-colored ring has a mass of 17.5 grams and a volume of 0.82 mL. What is the density of this ring?

max2010maxim [7]

Answer:

21 g/mL

Explanation:

To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.

4 0

3 years ago

Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,

ahrayia [7]

Answer:

Buffalo, NY was colder on February 9, 1934

Explanation:

Temperature is the measure of thermal energy present in a given substance. It is a physical property of a matter which can be measured on the Celsius scale (°C), Fahrenheit scale (°F), and Kelvin scale (K).

To convert Celsius (°C) to Fahrenheit (°F), we use the equation:

°F = 9/5 (°C) + 32

Given: On February 9, 1934,

Temperature in Buffalo, NY: T₁ = (- 29) °C

Temperature in Anchorage, AL: T₂ = 19 °F

To convert T₁ from °C to °F:

9/5 (- 29 °C) + 32 = (- 52.2) + 32 = (- 20.2) °F

So,

Temperature in Buffalo, NY: T₁ = (- 20.2) °F

Temperature in Anchorage, AL: T₂ = 19 °F

⇒ T₁ < T₂

Therefore, Buffalo, NY was colder on February 9, 1934.

4 0

3 years ago

4) The initial rate of the reaction between substances P and Q was measured in a series of

ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

$rate = k[P]^{2} [Q]$

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

$rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }$

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

$k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4$

Second row:

2. Rate value:

$rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1$

3.Third row:

$[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}$

4. Fourth row:

$[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}$

6 0

2 years ago

What would be the final volume of the new solution if the 0.2 m solution on the left were diluted to 0.04 m? ml quizley?

marin [14]

Missing question: volume of solution on the left is 10 mL.
V₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.
V₂(solution) = ?.
c₂(solution) = 0.04 M.
c₁ - original concentration of the solution, before it gets diluted.
c₂ - final concentration of the solution, after dilution.
V₁ - volume to be diluted.
V₂ - final volume after dilution.
c₁ · V₁ = c₂ · V₂.
10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M ÷ 0.04 M.
V₂(solution) = 50 mL.

3 0

3 years ago