Diagonal lines in the graph represent the temperature of the substance is __

The hot dogs you ate at the barbecue last week were 75% fat-free by weight, had 275 calories, and weighed 110 g. a) What percent

vovangra [49]

Answer:

a) 25%

b) 27.5 g

c) 90%

Explanation:

a) 75% fat-free by weight means 25% of the weight is made by fat.

b) 110 g ___ 100%

x ___ 25%

x = 27.5g

Each hot dog has 27.5g of fat.

c) 9 cal ___ 1 g fat

y ___ 27.5 g fat

y = 247.5 cal

275 cal ___ 100%

247.5 cal ___ z

z = 90%

90 % of the calories come from fat.

6 0

3 years ago

If 52500 J of heat is used to heat a 10200g block of metal and

Angelina_Jolie [31]

Answer:

45.95 Jkg^-1°C^-1

Explanation:

as specific heat capacity = heat energy / mass × delta

temperature

=52500/10.2×112

=45.95 Jkg^-1°C^-1

5 0

3 years ago

if the spin of one electron in an orbital is clockwise , what is the spin of the other electron in that orbital

nata0808 [166]

Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.

The electron will have anti-clockwise notation.

We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.

Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = $\frac{+}{-}$ $\frac{1}{2}$ .

The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:

1, 0, 0, + $\frac{1}{2}$ and 1, 0, 0, - $\frac{1}{2}$ .

Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.

6 0

3 years ago

What is the molecular mass of Br 2?

Kazeer [188]

Answer:

159.808 g/mol

3 0

3 years ago

Read 2 more answers

In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)

Veronika [31]

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

4 0

3 years ago

Diagonal lines in the graph represent the temperature of the substance is ___.