All categories

7 days ago

If apples are $.67/lb, what is the cost of 2.5 kg of apples?

Chemistry

2 answers:

VMariaS [2.6K]7 days ago

8 0

Answer:

The price will be $1.68

Explanation:

I arrived at this answer by multiplying

.67x or.67lb with 2.5, resulting in $1.68

lions [2.5K]7 days ago

7 0

Answer: $3.693

Explanation:

Since an apple is priced at $0.67 per lb

To determine the cost for 2.5kg, we first need to convert kg to lb

Given that 1kg = 2.205lb

2.5kg = (2.205×2.5)lb

2.5kg = 5.513lb

Since $0.67 applies to 1lb

$x = 5.513lb

If we cross multiply, we have;

x × 1 = 0.67×5.513

x = $3.693

You might be interested in

Which example best demonstrates stewardship of the atmosphere

KiRa [2612]

The question is incomplete. Nonetheless, I gave an example of atmospheric stewardship. First, it’s important to understand that stewardship implies a duty for humans to care for and preserve our environment, which includes the atmosphere. One way to demonstrate this is by changing how we travel; for example, opting to bike instead of using a car. This choice would help decrease greenhouse gas emissions since bicycles do not release harmful gases.

3 0

18 days ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

eduard [2409]

Details:

The equation to calculate work done is defined as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now we will insert the given values into the formula above to compute the work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is $3.68 \times 10^{-18} J$ .

7 0

19 days ago

A chalk company gets an order for 3000 boxes of chalk at the beginning of the school year, the largest order they have ever rece

VMariaS [2631]

The chalk production's efficiency is noted to be 82 %

To determine this efficiency, we utilize the formula,

% Efficiency $=\frac{Actual yield (g)}{Theoretical yield (g)} *100$

Finding the actual yield based on % efficiency and the theoretical yield:

82 % $=\frac{Actual yield}{400,000 g} * 100$

Actual yield = $\frac{82 *400000 g}{100} = 32,800 g$

Each box contains 145 grams of chalk

Total boxes = $32800 g *\frac{1 box}{145 g} = 2262 boxes$

The chalk box company aims to produce 3000 boxes. However, with 82 % efficiency, they can only produce 2262 boxes, thus falling short of their target.

8 0

27 days ago

The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

KiRa [2612]

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

7 0

24 days ago

A piece of iron (C=0.449 J/g°C) and a piece of gold (C=0.128 J/g°C) have identical masses. If the iron has an initial temperatur

lorasvet [2450]

The true statement is B. With identical masses for both metals, the final temperature of the two will be more aligned with 498 K rather than 298 K, as iron's specific heat capacity is significantly greater than that of gold's.

4 0

19 days ago